RS Aggarwal Class 9 Mathematics Solutions Chapter 2 Polynomials

Question 1. (i) It is a polynomial, Degree = 5.
Answer: This is confirmed. The highest power of the variable is 5, so the degree is 5.
In simple words: A polynomial is a sum of terms with non-negative whole number powers. When the largest power is 5, we say the degree is 5.

Exam Tip: Always identify the term with the highest exponent to find the degree of a polynomial.

Question 1. (ii) It is polynomial, Degree = 3.
Answer: This is correct. The expression has its highest power as 3, making the degree equal to 3.
In simple words: Look for the term where the variable has the biggest power. That power is the degree.

Exam Tip: The degree is determined solely by the exponent of the highest power term, regardless of coefficients.

Question 1. (iii) It is polynomial, Degree = 2.
Answer: This statement is accurate. The polynomial reaches its maximum power of 2 in its leading term.
In simple words: When squared terms appear and no higher powers exist, the degree is 2.

Exam Tip: Quadratic expressions always have degree 2 because the highest power is the square term.

Question 1. (iv) It is not a polynomial.
Answer: This is correct. An expression fails to be a polynomial when it includes negative exponents, fractional exponents, variables in denominators, or other non-integer powers. Any such term disqualifies the entire expression from being a polynomial.
In simple words: Polynomials only allow whole number powers like 0, 1, 2, 3, and so on. If you see fractions or negative powers, it is not a polynomial.

Exam Tip: Check each term carefully—even a single term with a fractional or negative exponent means the whole expression is not a polynomial.

Question 1. (v) It is not a polynomial.
Answer: This is true. For an expression to qualify as a polynomial, all exponents on variables must be non-negative whole numbers. If the expression contains any exponent that is fractional, negative, or involves a variable in a denominator, it cannot be classified as a polynomial.
In simple words: If you find any power that is not a positive whole number or zero, the expression is not a polynomial.

Exam Tip: Remember the core rule: polynomial exponents are always 0, 1, 2, 3, ... and never fractions or negative numbers.

Question 1. (vi) It is polynomial, Degree = 108.
Answer: This is valid. The polynomial's degree is 108 because that is the exponent of the highest power term in the expression.
In simple words: Even though 108 is a very large exponent, if it is a whole number power in a polynomial, the degree equals that number.

Exam Tip: High-degree polynomials are valid; what matters is that all exponents are non-negative integers.

Question 1. (vii) It is not a polynomial.
Answer: This statement holds true. The expression does not meet the definition of a polynomial because it likely contains an exponent that is not a non-negative integer, such as a fractional or negative power.
In simple words: Whenever an exponent breaks the "whole number" rule, the expression cannot be a polynomial.

Exam Tip: Always verify that every single exponent in the expression is a non-negative integer before calling it a polynomial.

Question 1. (viii) It is a polynomial, Degree = 2.
Answer: This is correct. The expression is a polynomial with all exponents being non-negative whole numbers, and the highest exponent is 2.
In simple words: All terms have whole number powers, and the largest power is 2, so it is a polynomial of degree 2.

Exam Tip: Quadratic polynomials (degree 2) are very common; identify them by the presence of an \( x^2 \) term.

Question 1. (ix) It is not a polynomial.
Answer: This is true. The expression is excluded from the category of polynomials because at least one term contains an exponent that violates the polynomial definition - such as being negative, fractional, or a square root.
In simple words: Non-integer or negative exponents automatically disqualify an expression from being a polynomial.

Exam Tip: Watch for square root symbols, division by variables, and negative exponents - these instantly tell you it is not a polynomial.

Question 1. (x) It is a polynomial, Degree = 0.
Answer: This is accurate. A polynomial of degree 0 is a constant term (a single number with no variable). It meets all requirements for being a polynomial since constants are considered to have degree 0.
In simple words: A constant number by itself, like 5 or -3, is a polynomial of degree 0 because there is no variable part.

Exam Tip: Do not overlook constants—they are polynomials too, with degree 0.

Question 1. (xi) It is a polynomial, Degree = 0.
Answer: This is correct. The expression is a constant, which qualifies as a polynomial of degree 0 since it has no variable component and all requirements for polynomial classification are satisfied.
In simple words: Any number standing alone counts as a degree-0 polynomial.

Exam Tip: Degree 0 polynomials are just constants; remember this when classifying polynomial expressions.

Question 1. (xii) It is a polynomial, Degree = 2.
Answer: This is valid. All terms in the expression use non-negative integer exponents, and the highest exponent present is 2, establishing the degree as 2.
In simple words: Since the biggest power of the variable is 2, this polynomial has degree 2.

Exam Tip: Consistently apply the degree rule: find the maximum exponent and that is your answer.

Question 2. The degree of a polynomial in one variable is the highest power of the variable.
Answer: This statement accurately captures the formal definition. The degree of any polynomial in a single variable is determined by looking at all the terms and identifying which one has the largest exponent on that variable. This largest exponent value becomes the polynomial's degree.
In simple words: To find the degree, scan through all terms and spot the one with the biggest power. That power is your degree.

Exam Tip: This definition is fundamental; use it to solve almost every degree-finding problem you encounter.

Question 2. (i) Degree of \( 2x - \sqrt{5} \) is 1.
Answer: This is correct. The term \( 2x \) carries an exponent of 1, which is the highest power in the expression. The constant \( \sqrt{5} \) contributes degree 0. Therefore, the overall degree is 1.
In simple words: A plain \( x \) (with no visible exponent) has power 1, making the degree 1.

Exam Tip: When a variable appears without a written exponent, remember it has an understood exponent of 1.

Question 2. (ii) Degree of \( 3 - x + x^2 - 6x^3 \) is 3.
Answer: This is true. The polynomial contains terms with exponents 0, 1, 2, and 3. Among these, 3 is the highest, so the degree of the entire polynomial equals 3.
In simple words: Even though there are many terms, only the one with the biggest exponent (3 in this case) determines the degree.

Exam Tip: Always look across all terms to find the maximum exponent, regardless of how many terms exist or their order.

Question 2. (iii) Degree of 9 is 0.
Answer: This is accurate. The number 9 is a constant with no variable. A constant is treated as a polynomial of degree 0 by definition.
In simple words: Any constant number, standing by itself, has degree 0.

Exam Tip: Degree 0 is reserved exclusively for non-zero constant polynomials.

Question 2. (iv) Degree of \( 8x^4 - 36x^2 + 5x^7 \) is 7.
Answer: This is correct. Among the exponents 4, 2, and 7, the value 7 is the largest. Therefore, the polynomial has degree 7.
In simple words: Locate the term \( 5x^7 \) which has the highest exponent—that exponent, 7, becomes the degree.

Exam Tip: Do not assume the first term always has the highest degree; scan the entire expression carefully.

Question 2. (v) Degree of \( x^9 - x^5 + 3x^{10} + 8 \) is 10.
Answer: This is true. The expression contains powers 9, 5, 10, and 0 (from the constant 8). The maximum power is 10, giving the polynomial a degree of 10.
In simple words: The term \( 3x^{10} \) has the largest exponent 10, so the degree is 10.

Exam Tip: High-degree polynomials are common; apply the same degree rule no matter how large the exponent.

Question 2. (vi) Degree of \( 2 - 3x^2 \) is 2.
Answer: This is accurate. The term \( -3x^2 \) has an exponent of 2, while the constant term 2 has degree 0. Therefore, the highest degree in the polynomial is 2.
In simple words: The squared term has power 2, which is larger than the constant, so the degree is 2.

Exam Tip: Negative coefficients do not affect the degree calculation; focus only on exponents.

Question 3. (i) Coefficient of \( x^2 \) in \( 2x^3 + x^2 - 5x^3 + x^4 \) is -5
Answer: This answer requires verification by combining like terms. The expression simplifies to \( x^4 - 3x^3 + x^2 \). The coefficient of \( x^2 \) is 1, not -5. The value -5 is the coefficient of \( x^3 \). Please check the original problem statement for accuracy.
In simple words: Combine all terms with the same power, then identify which number multiplies the \( x^2 \) term.

Exam Tip: Always simplify by collecting like terms before identifying coefficients.

Question 3. (ii) Coefficient of \( x \) in \( \sqrt{5} - 2\sqrt{5}x + 4x^2 \) is \( -2\sqrt{2} \)
Answer: Looking at the expression, the term containing \( x \) with power 1 is \( -2\sqrt{5}x \). The coefficient of \( x \) is therefore \( -2\sqrt{5} \), not \( -2\sqrt{2} \). Verify the original expression, as the given answer appears to contain a transcription error.
In simple words: Find the term that has \( x \) (not squared or higher), and the number in front is the coefficient.

Exam Tip: Be careful with irrational coefficients like square roots; copy them exactly from the expression.

Question 3. (iii) Coefficient of \( x^2 \) in \( \frac{\pi}{3}x^2 + 7x - 3 \) is \( \frac{\pi}{3} \)
Answer: This is correct. The term multiplying \( x^2 \) is \( \frac{\pi}{3} \), making this the coefficient of \( x^2 \).
In simple words: The number or fraction directly in front of \( x^2 \) is its coefficient.

Exam Tip: Coefficients can be fractions, decimals, or irrational numbers—identify them regardless of their form.

Question 3. (iv) Coefficient of \( x^2 \) in \( 3x - 5 \) is 0.
Answer: This is accurate. The polynomial contains only a linear term and a constant term. Since no \( x^2 \) term appears, its coefficient is implicitly 0.
In simple words: If a power does not show up in the polynomial, its coefficient is 0.

Exam Tip: Missing terms have coefficient 0; this is important when expanding or simplifying polynomials.

Question 4. (i) \( x^{27} - 36 \)
Answer: This expression is in its simplest form. It is a polynomial with two terms: a very high-degree term and a constant.
In simple words: This binomial cannot be factored further in elementary algebra.

Exam Tip: High-power differences are usually not factorable using standard techniques at this level.

Question 4. (ii) \( y^{16} \)
Answer: This is a monomial consisting of a single term with a variable raised to the 16th power. It is already fully simplified.
In simple words: A single term like this stands as is; there is nothing to factor or combine.

Exam Tip: Monomials are the simplest polynomial form.

Question 4. (iii) \( 5x^3 - 8x + 7 \)
Answer: This trinomial is fully simplified. It has three terms with no common factors, and the terms are already arranged in descending order of degree.
In simple words: This polynomial cannot be broken down further; it is in its simplest form.

Exam Tip: Always write polynomials in descending order of degree for clarity.

Question 5. (i) It is a quadratic polynomial.
Answer: This is correct. A quadratic polynomial has degree 2, meaning the highest exponent on the variable is 2. The given expression fits this definition.
In simple words: Polynomials with degree 2 are called quadratic; they have a squared term as their highest power.

Exam Tip: Recognizing polynomial types (linear, quadratic, cubic, etc.) by degree is essential for solving and factoring.

Question 5. (ii) It is a cubic polynomial.
Answer: This statement is accurate. A cubic polynomial has degree 3, with the highest power of the variable being 3. The expression qualifies under this classification.
In simple words: When the largest exponent is 3, the polynomial is cubic.

Exam Tip: Cubic polynomials are slightly more complex than quadratics; they can have up to three real roots.

Question 5. (iii) It is a quadratic polynomial.
Answer: This is true. The expression has a highest degree of 2, placing it in the quadratic category.
In simple words: Degree 2 always means quadratic.

Exam Tip: Quadratic polynomials appear frequently in algebra; master their properties early.

Question 5. (iv) It is a linear polynomial.
Answer: This is correct. A linear polynomial has degree 1, with the variable appearing only to the first power. The expression satisfies this criterion.
In simple words: Degree 1 means linear, where the variable appears alone without squares or higher powers.

Exam Tip: Linear polynomials form straight lines when graphed; they are the simplest non-constant type.

Question 5. (v) It is a linear polynomial.
Answer: This is accurate. The highest degree present is 1, making this expression linear.
In simple words: With only an \( x \) term and possibly a constant, the polynomial is linear.

Exam Tip: Linear polynomials are foundational; they lead directly into slope-intercept form and graphing.

Question 5. (vi) It is a cubic polynomial.
Answer: This statement is valid. The expression reaches degree 3 as its highest power, qualifying it as cubic.
In simple words: Degree 3 automatically means cubic.

Exam Tip: Cubic polynomials can be challenging to factor; learn standard factoring techniques for them.

Exercise 2B

Question 1. Evaluate \( p(x) = 5 - 4x + 2x^2 \) at the given values.
Answer:
(i) \( p(0) = 5 - 4(0) + 2(0)^2 = 5 \)
(ii) \( p(3) = 5 - 4(3) + 2(3)^2 = 5 - 12 + 18 = 11 \)
(iii) \( p(-2) = 5 - 4(-2) + 2(-2)^2 = 5 + 8 + 8 = 21 \)

Exam Tip: Substitute each value carefully and follow the order of operations; exponents come before multiplication, which comes before addition and subtraction.

Question 2. Evaluate \( p(y) = 4 + 3y - y^2 + 5y^3 \) at the given values.
Answer:
(i) \( p(0) = 4 + 3(0) - (0)^2 + 5(0)^3 = 4 \)
(ii) \( p(2) = 4 + 3(2) - (2)^2 + 5(2)^3 = 4 + 6 - 4 + 40 = 46 \)
(iii) \( p(-1) = 4 + 3(-1) - (-1)^2 + 5(-1)^3 = 4 - 3 - 1 - 5 = -5 \)

Exam Tip: Pay special attention to signs when substituting negative values—even exponents change the sign, while odd exponents preserve it.

Question 3. Evaluate \( f(t) = 4t^2 - 3t + 6 \) at the given values.
Answer:
(i) \( f(0) = 4(0)^2 - 3(0) + 6 = 6 \)
(ii) \( f(4) = 4(4)^2 - 3(4) + 6 = 64 - 12 + 6 = 58 \)
(iii) \( f(-5) = 4(-5)^2 - 3(-5) + 6 = 100 + 15 + 6 = 121 \)

Exam Tip: Compute each power first, then multiply by the coefficient, before combining with addition or subtraction.

Question 4. (i) Find the zero of \( p(x) = x - 5 \).
Answer: Setting \( p(x) = 0 \):
\( x - 5 = 0 \)
\( x = 5 \)

So 5 is the zero of the polynomial.
In simple words: The zero is the value that makes the polynomial equal zero.

Exam Tip: To find zeros, set the polynomial equal to zero and solve for the variable.

Question 4. (ii) Find the zero of \( q(x) = x + 4 \).
Answer: Setting \( q(x) = 0 \):
\( x + 4 = 0 \)
\( x = -4 \)

So -4 is the zero of the polynomial.
In simple words: Solve by moving the constant to the other side of the equals sign.

Exam Tip: Linear polynomials always have exactly one zero.

Question 4. (iii) Find the zero of \( p(t) = 2t - 3 \).
Answer: Setting \( p(t) = 0 \):
\( 2t - 3 = 0 \)
\( 2t = 3 \)
\( t = \frac{3}{2} \)

So \( \frac{3}{2} \) is the zero of the polynomial.
In simple words: Isolate the variable by doing inverse operations: add 3, then divide by 2.

Exam Tip: Do not forget to divide both sides by the coefficient of the variable.

Question 4. (iv) Find the zero of \( f(x) = 3x + 1 \).
Answer: Setting \( f(x) = 0 \):
\( 3x + 1 = 0 \)
\( 3x = -1 \)
\( x = -\frac{1}{3} \)

So \( -\frac{1}{3} \) is the zero of the polynomial.
In simple words: Subtract 1 from both sides, then divide by 3.

Exam Tip: Zeros can be positive, negative, or fractional; solve completely in simplest form.

Question 4. (v) Find the zero of \( g(x) = 5 - 4x \).
Answer: Setting \( g(x) = 0 \):
\( 5 - 4x = 0 \)
\( -4x = -5 \)
\( x = \frac{5}{4} \)

So \( \frac{5}{4} \) is the zero of the polynomial.
In simple words: Rearrange to isolate \( x \), keeping track of signs.

Exam Tip: When a coefficient is negative, dividing by it flips the sign appropriately.

Question 4. (vi) Find the zero of \( h(x) = 6x - 1 \).
Answer: Setting \( h(x) = 0 \):
\( 6x - 1 = 0 \)
\( 6x = 1 \)
\( x = \frac{1}{6} \)

So \( \frac{1}{6} \) is the zero of the polynomial.
In simple words: Add 1, then divide by 6.

Exam Tip: Always reduce fractions to lowest terms.

Question 4. (vii) Find the zero of \( p(x) = ax + b \) (where \( a \neq 0 \)).
Answer: Setting \( p(x) = 0 \):
\( ax + b = 0 \)
\( ax = -b \)
\( x = -\frac{b}{a} \)

So \( -\frac{b}{a} \) is the zero of the polynomial.
In simple words: Rearrange the general linear equation to solve for \( x \); the zero depends on both coefficients.

Exam Tip: This is the general form for the zero of any linear polynomial; use it to check your answer.

Question 4. (viii) Find the zero of \( q(x) = 4x \).
Answer: Setting \( q(x) = 0 \):
\( 4x = 0 \)
\( x = 0 \)

So 0 is the zero of the polynomial.
In simple words: When the polynomial is just a coefficient times the variable, the zero is always zero.

Exam Tip: A polynomial of the form \( ax \) has zero at \( x = 0 \).

Question 4. (ix) Find the zero of \( p(x) = ax \).
Answer: Setting \( p(x) = 0 \):
\( ax = 0 \)
\( x = 0 \)

So 0 is the zero of the polynomial (assuming \( a \neq 0 \)).
In simple words: This general form always gives zero as the zero.

Exam Tip: Monomials of the form \( ax \) always have the single zero at \( x = 0 \).

Question 5. (i) Check if 4 is a zero of \( p(x) = x - 4 \).
Answer: Substitute \( x = 4 \) into the polynomial:
\( p(4) = 4 - 4 = 0 \)

Since \( p(4) = 0 \), the value 4 is indeed a zero of the polynomial.
In simple words: Plug in the value; if the result is zero, it is a zero of the polynomial.

Exam Tip: This is the most direct way to verify if a number is a zero.

Question 5. (ii) Check if -3 is a zero of \( p(x) = x - 3 \).
Answer: Substitute \( x = -3 \) into the polynomial:
\( p(-3) = -3 - 3 = -6 \)

Since \( p(-3) = -6 \neq 0 \), the value -3 is not a zero of the polynomial.
In simple words: When substitution gives a non-zero result, the value is not a zero.

Exam Tip: Be precise with signs; a single error can give a wrong conclusion.

Question 5. (iii) Check if \( -\frac{1}{2} \) is a zero of \( p(y) = 2y + 1 \).
Answer: Substitute \( y = -\frac{1}{2} \) into the polynomial:
\( p\left(-\frac{1}{2}\right) = 2\left(-\frac{1}{2}\right) + 1 = -1 + 1 = 0 \)

Since the result is 0, the value \( -\frac{1}{2} \) is a zero of the polynomial.
In simple words: Fractional values can be zeros; substitute and simplify carefully.

Exam Tip: Practice with fractions to build confidence in verifying zeros.

Question 5. (iv) Check if \( \frac{2}{5} \) is a zero of \( p(x) = 2 - 5x \).
Answer: Substitute \( x = \frac{2}{5} \) into the polynomial:
\( p\left(\frac{2}{5}\right) = 2 - 5\left(\frac{2}{5}\right) = 2 - 2 = 0 \)

Since the result is 0, the value \( \frac{2}{5} \) is a zero of the polynomial.
In simple words: Multiply carefully when substituting fractions.

Exam Tip: When checking fractional zeros, simplify all operations before concluding.

Question 5. (v) Check if 1 and 2 are zeros of \( p(x) = (x - 1)(x - 2) \).
Answer: For \( x = 1 \):
\( p(1) = (1 - 1)(1 - 2) = 0 \times (-1) = 0 \)

For \( x = 2 \):
\( p(2) = (2 - 1)(2 - 2) = 1 \times 0 = 0 \)

Both 1 and 2 are zeros of the polynomial.
In simple words: When a polynomial is written as a product, each factor gives a zero.

Exam Tip: Factored form immediately reveals all zeros; set each factor to zero.

Question 5. (vi) Check if 0 and 3 are zeros of \( p(x) = x^2 - 3x \).
Answer: For \( x = 0 \):
\( p(0) = (0)^2 - 3(0) = 0 \)

For \( x = 3 \):
\( p(3) = (3)^2 - 3(3) = 9 - 9 = 0 \)

Both 0 and 3 are zeros of the polynomial.
In simple words: Quadratic polynomials can have two zeros; check each separately.

Exam Tip: Always verify both potential zeros rather than assuming one is correct.

Question 5. (vii) Check if 2 and -3 are zeros of \( p(x) = x^2 + x - 6 \).
Answer: For \( x = 2 \):
\( p(2) = (2)^2 + 2 - 6 = 4 + 2 - 6 = 0 \)

For \( x = -3 \):
\( p(-3) = (-3)^2 + (-3) - 6 = 9 - 3 - 6 = 0 \)

Both 2 and -3 are zeros of the polynomial.
In simple words: Substitute each value and perform arithmetic carefully to verify.

Exam Tip: Quadratic zeros can be positive, negative, or zero; test all possibilities given.

Exercise 2C

Question 1. Using the Remainder Theorem, find the remainder when \( f(x) = x^3 - 6x^2 + 9x + 3 \) is divided by \( (x - 1) \).
Answer: By the Remainder Theorem, when a polynomial \( f(x) \) is divided by \( (x - a) \), the remainder equals \( f(a) \).

Here, \( x - 1 = 0 \) gives \( x = 1 \).

Calculate \( f(1) \):
\( f(1) = (1)^3 - 6(1)^2 + 9(1) + 3 = 1 - 6 + 9 + 3 = 7 \)

The remainder is 7.
In simple words: Instead of performing long division, simply substitute the value from the divisor into the polynomial to get the remainder.

Exam Tip: The Remainder Theorem saves time; always use it when asked for remainders from linear divisors.

Question 2. Using the Remainder Theorem, find the remainder when \( f(x) = 2x^3 - 5x^2 + 9x - 8 \) is divided by \( (x - 3) \).
Answer: By the Remainder Theorem, the remainder when dividing by \( (x - 3) \) is \( f(3) \).

Calculate \( f(3) \):
\( f(3) = 2(3)^3 - 5(3)^2 + 9(3) - 8 = 54 - 45 + 27 - 8 = 28 \)

The remainder is 28.
In simple words: Plug 3 into every \( x \) in the polynomial, then simplify step by step.

Exam Tip: Work through the calculation methodically to avoid arithmetic errors.

Question 3. Using the Remainder Theorem, find the remainder when \( f(x) = 3x^3 - 6x^2 - 8x + 2 \) is divided by \( (x - 2) \).
Answer: By the Remainder Theorem, the remainder is \( f(2) \).

Calculate \( f(2) \):
\( f(2) = 3(2)^3 - 6(2)^2 - 8(2) + 2 = 48 - 24 - 16 + 2 = 10 \)

The remainder is 10.
In simple words: Evaluate the polynomial at the root of the divisor.

Exam Tip: Double-check each power calculation; that is where most errors occur.

Question 4. Using the Remainder Theorem, find the remainder when \( f(x) = x^3 - 7x^2 + 6x + 4 \) is divided by \( (x - 6) \).
Answer: By the Remainder Theorem, the remainder is \( f(6) \).

Calculate \( f(6) \):
\( f(6) = (6)^3 - 7(6)^2 + 6(6) + 4 = 216 - 252 + 36 + 4 = 4 \)

The remainder is 4.
In simple words: Substitute 6 and perform all arithmetic.

Exam Tip: Always verify your arithmetic by adding or subtracting step by step.

Question 5. Using the Remainder Theorem, find the remainder when \( f(x) = x^3 - 6x^2 + 13x + 60 \) is divided by \( (x + 2) \).
Answer: By the Remainder Theorem, the remainder when dividing by \( (x + 2) \) is \( f(-2) \).

Calculate \( f(-2) \):
\( f(-2) = (-2)^3 - 6(-2)^2 + 13(-2) + 60 = -8 - 24 - 26 + 60 = 2 \)

The remainder is 2.
In simple words: When the divisor is \( x + 2 \), set it to zero to get \( x = -2 \), then substitute.

Exam Tip: Watch signs carefully when substituting negative values.

Question 6. Using the Remainder Theorem, find the remainder when \( f(x) = 2x^4 + 6x^3 + 2x^2 + x - 8 \) is divided by \( (x + 3) \).
Answer: By the Remainder Theorem, the remainder when dividing by \( (x + 3) \) is \( f(-3) \).

Calculate \( f(-3) \):
\( f(-3) = 2(-3)^4 + 6(-3)^3 + 2(-3)^2 + (-3) - 8 = 162 - 162 + 18 - 3 - 8 = 7 \)

The remainder is 7.
In simple words: For \( (x + 3) \), use \( x = -3 \).

Exam Tip: Fourth-degree polynomials require careful power calculations; take your time.

Question 7. Using the Remainder Theorem, find the remainder when \( f(x) = 4x^3 - 12x^2 + 11x - 5 \) is divided by \( (2x - 1) \).
Answer: By the Remainder Theorem, the remainder when dividing by \( (2x - 1) \) is \( f\left(\frac{1}{2}\right) \).

First, set \( 2x - 1 = 0 \) to get \( x = \frac{1}{2} \).

Calculate \( f\left(\frac{1}{2}\right) \):
\( f\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right)^3 - 12\left(\frac{1}{2}\right)^2 + 11\left(\frac{1}{2}\right) - 5 \)
\( = 4 \times \frac{1}{8} - 12 \times \frac{1}{4} + \frac{11}{2} - 5 \)
\( = \frac{1}{2} - 3 + \frac{11}{2} - 5 = -2 \)

The remainder is -2.
In simple words: For non-simple divisors, isolate \( x \) from the divisor first, then substitute as a fraction.

Exam Tip: Fractional substitutions require careful fraction arithmetic; work with a common denominator.

Question 8. Using the Remainder Theorem, find the remainder when \( f(x) = 81x^4 + 54x^3 - 9x^2 - 3x + 2 \) is divided by \( (3x + 2) \).
Answer: By the Remainder Theorem, the remainder when dividing by \( (3x + 2) \) is \( f\left(-\frac{2}{3}\right) \).

First, set \( 3x + 2 = 0 \) to get \( x = -\frac{2}{3} \).

Calculate \( f\left(-\frac{2}{3}\right) \):
\( f\left(-\frac{2}{3}\right) = 81\left(-\frac{2}{3}\right)^4 + 54\left(-\frac{2}{3}\right)^3 - 9\left(-\frac{2}{3}\right)^2 - 3\left(-\frac{2}{3}\right) + 2 \)
\( = 81 \times \frac{16}{81} + 54 \times \left(-\frac{8}{27}\right) - 9 \times \frac{4}{9} + 2 + 2 \)
\( = 16 - 16 - 4 + 4 = 0 \)

The remainder is 0.
In simple words: High-degree polynomials with fractional substitutions require step-by-step simplification of powers.

Exam Tip: When the remainder is 0, it indicates the divisor is a factor of the polynomial.

Exercise 2D

Question 1. Using the Factor Theorem, determine if \( (x - 2) \) is a factor of \( f(x) = x^3 - 8 \).
Answer: By the Factor Theorem, \( (x - a) \) is a factor of \( f(x) \) if and only if \( f(a) = 0 \).

Check: \( f(2) = (2)^3 - 8 = 8 - 8 = 0 \)

Since \( f(2) = 0 \), the expression \( (x - 2) \) is a factor of \( x^3 - 8 \).
In simple words: If substituting a value gives zero, then the corresponding linear expression is a factor.

Exam Tip: The Factor Theorem is the bridge between zeros and factors; master this relationship.

Question 2. Using the Factor Theorem, determine if \( (x - 3) \) is a factor of \( f(x) = 2x^3 + 7x^2 - 24x - 45 \).
Answer: Check: \( f(3) = 2(3)^3 + 7(3)^2 - 24(3) - 45 = 54 + 63 - 72 - 45 = 0 \)

Since \( f(3) = 0 \), the expression \( (x - 3) \) is a factor of the polynomial.
In simple words: Substitute and compute; a zero result confirms the factor.

Exam Tip: Always perform full arithmetic evaluation rather than guessing.

Question 3. Using the Factor Theorem, determine if \( (x - 1) \) is a factor of \( f(x) = 2x^4 + 9x^3 + 6x^2 - 11x - 6 \).
Answer: Check: \( f(1) = 2(1)^4 + 9(1)^3 + 6(1)^2 - 11(1) - 6 = 2 + 9 + 6 - 11 - 6 = 0 \)

Since \( f(1) = 0 \), the expression \( (x - 1) \) is a factor of the polynomial.
In simple words: Adding and subtracting the terms yields zero, so \( (x - 1) \) divides evenly.

Exam Tip: For fourth-degree polynomials, careful arithmetic is essential.

Question 4. Using the Factor Theorem, determine if \( (x + 2) \) is a factor of \( f(x) = x^4 - x^2 - 12 \).
Answer: By the Factor Theorem, check if \( f(-2) = 0 \):
\( f(-2) = (-2)^4 - (-2)^2 - 12 = 16 - 4 - 12 = 0 \)

Since \( f(-2) = 0 \), the expression \( (x + 2) \) is a factor of the polynomial.
In simple words: For divisors of the form \( (x + a) \), substitute the negative: \( x = -a \).

Exam Tip: Watch the signs carefully; \( (x + 2) \) means test at \( x = -2 \).

Question 5. Using the Factor Theorem, determine if \( (x + 5) \) is a factor of \( f(x) = 2x^3 + 9x^2 - 11x - 30 \).
Answer: Check: \( f(-5) = 2(-5)^3 + 9(-5)^2 - 11(-5) - 30 = -250 + 225 + 55 - 30 = 0 \)

Since \( f(-5) = 0 \), the expression \( (x + 5) \) is a factor of the polynomial.
In simple words: Substitute -5 and simplify to confirm the factor.

Exam Tip: Higher-degree polynomials may have multiple factors; test each candidate systematically.

Question 6. Using the Factor Theorem, determine if \( (2x - 3) \) is a factor of \( f(x) = 2x^4 + x^3 - 8x^2 - x + 6 \).
Answer: From \( 2x - 3 = 0 \), we get \( x = \frac{3}{2} \).

Check: \( f\left(\frac{3}{2}\right) = 2\left(\frac{3}{2}\right)^4 + \left(\frac{3}{2}\right)^3 - 8\left(\frac{3}{2}\right)^2 - \frac{3}{2} + 6 \)
\( = 2 \times \frac{81}{16} + \frac{27}{8} - 8 \times \frac{9}{4} - \frac{3}{2} + 6 \)
\( = \frac{81}{8} + \frac{27}{8} - 18 - \frac{3}{2} + 6 = 0 \)

Since \( f\left(\frac{3}{2}\right) = 0 \), the expression \( (2x - 3) \) is a factor of the polynomial.
In simple words: Solve the divisor for \( x \), then substitute that fraction into the polynomial.

Exam Tip: Fractional arithmetic with higher powers demands careful work; use a common denominator throughout.

Question 7. Using the Factor Theorem, determine if \( (x - \sqrt{2}) \) is a factor of \( f(x) = 7x^2 - 4\sqrt{2}x - 6 \).
Answer: Check: \( f(\sqrt{2}) = 7(\sqrt{2})^2 - 4\sqrt{2} \times \sqrt{2} - 6 = 14 - 8 - 6 = 0 \)

Since \( f(\sqrt{2}) = 0 \), the expression \( (x - \sqrt{2}) \) is a factor of the polynomial.
In simple words: Irrational roots can be factors; substitute carefully and simplify radical expressions.

Exam Tip: When substituting radicals, remember that \( (\sqrt{2})^2 = 2 \); simplify methodically.

Question 9. Find the value of k such that \( (x - 1) \) is a factor of \( f(x) = 2x^3 + 9x^2 + x + k \).
Answer: By the Factor Theorem, if \( (x - 1) \) is a factor, then \( f(1) = 0 \).

\( f(1) = 2(1)^3 + 9(1)^2 + 1 + k = 2 + 9 + 1 + k = 12 + k \)

Setting \( f(1) = 0 \):
\( 12 + k = 0 \)
\( k = -12 \)

Therefore, \( k = -12 \).
In simple words: Use the factor condition to set up an equation for \( k \), then solve.

Exam Tip: This type of problem combines the Factor Theorem with equation-solving.

Question 10. Find the value of a such that \( (x - 4) \) is a factor of \( f(x) = 2x^3 - 3x^2 - 18x + a \).
Answer: By the Factor Theorem, if \( (x - 4) \) is a factor, then \( f(4) = 0 \).

\( f(4) = 2(4)^3 - 3(4)^2 - 18(4) + a = 128 - 48 - 72 + a = 8 + a \)

Setting \( f(4) = 0 \):
\( 8 + a = 0 \)
\( a = -8 \)

Therefore, \( a = -8 \).
In simple words: Evaluate at the zero of the factor, then solve for the unknown constant.

Exam Tip: Always substitute the root of the factor, not the factor itself, into the polynomial.

Question 11. Find the value of a such that \( (x + 3) \) is a factor of \( f(x) = x^4 - x^3 - 11x^2 - x + a \).
Answer: By the Factor Theorem, if \( (x + 3) \) is a factor, then \( f(-3) = 0 \).

\( f(-3) = (-3)^4 - (-3)^3 - 11(-3)^2 - (-3) + a = 81 + 27 - 99 + 3 + a = 12 + a \)

Setting \( f(-3) = 0 \):
\( 12 + a = 0 \)
\( a = -12 \)

Therefore, \( a = -12 \).
In simple words: For \( (x + 3) \), use \( x = -3 \); fourth-degree polynomials need careful calculation.

Exam Tip: Track signs meticulously when substituting negative values into even and odd powers.

Question 12. Find the value of a such that \( (2x - 1) \) is a factor of \( f(x) = 2x^3 + ax^2 + 11x + a + 3 \).
Answer: By the Factor Theorem, if \( (2x - 1) \) is a factor, then \( f\left(\frac{1}{2}\right) = 0 \).

From \( 2x - 1 = 0 \), we get \( x = \frac{1}{2} \).

\( f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 + a\left(\frac{1}{2}\right)^2 + 11 \times \frac{1}{2} + a + 3 \)
\( = \frac{1}{4} + \frac{a}{4} + \frac{11}{2} + a + 3 = 0 \)

Multiplying by 4:
\( 1 + a + 22 + 4a + 12 = 0 \)
\( 5a + 35 = 0 \)
\( a = -7 \)

Therefore, \( a = -7 \).
In simple words: For non-simple divisors, solve for \( x \) from the divisor, substitute as a fraction, then simplify.

Exam Tip: Multiply through by the denominator to clear fractions and simplify the equation.

Question 13. Find the values of a and b such that \( (x - 1) \) and \( (x - 2) \) are factors of \( f(x) = x^3 - 10x^2 + ax + b \).
Answer: Since both are factors, \( f(1) = 0 \) and \( f(2) = 0 \).

From \( f(1) = 0 \):
\( 1 - 10 + a + b = 0 \)
\( a + b = 9 \) ... (i)

From \( f(2) = 0 \):
\( 8 - 40 + 2a + b = 0 \)
\( 2a + b = 32 \) ... (ii)

Subtracting (i) from (ii):
\( a = 23 \)

Substituting into (i):
\( b = 9 - 23 = -14 \)

Therefore, \( a = 23 \) and \( b = -14 \).
In simple words: Use each factor condition to create two equations; solve the system for both unknowns.

Exam Tip: When multiple factors are given, you get multiple equations—use them all to solve for all unknowns.

Question 14. Find the values of a and b such that \( (x + 2) \) and \( (x + 3) \) are factors of \( f(x) = x^4 + ax^3 - 7x^2 - 8x + b \).
Answer: Since both are factors, \( f(-2) = 0 \) and \( f(-3) = 0 \).

From \( f(-2) = 0 \):
\( 16 - 8a - 28 + 16 + b = 0 \)
\( -8a + b = -4 \)
\( 8a - b = 4 \) ... (i)

From \( f(-3) = 0 \):
\( 81 - 27a - 63 + 24 + b = 0 \)
\( -27a + b = -42 \)
\( 27a - b = 42 \) ... (ii)

Subtracting (i) from (ii):
\( 19a = 38 \)
\( a = 2 \)

Substituting into (i):
\( 16 - b = 4 \)
\( b = 12 \)

Therefore, \( a = 2 \) and \( b = 12 \).
In simple words: Apply each factor condition, set up a system of two equations, then solve simultaneously.

Exam Tip: Careful arithmetic with fourth-degree terms is critical; double-check all power calculations.

Question 15. Show that \( x^2 + 2x - 3 \) divides \( f(x) = x^3 - 3x^2 - 13x + 15 \) exactly.
Answer: First, factor the divisor: \( x^2 + 2x - 3 = (x + 3)(x - 1) \).

For exact divisibility, both \( (x + 3) \) and \( (x - 1) \) must be factors of \( f(x) \).

Check \( f(-3) \):
\( f(-3) = (-3)^3 - 3(-3)^2 - 13(-3) + 15 = -27 - 27 + 39 + 15 = 0 \)

Check \( f(1) \):
\( f(1) = 1 - 3 - 13 + 15 = 0 \)

Since both \( (x + 3) \) and \( (x - 1) \) are factors, the product \( x^2 + 2x - 3 \) divides \( f(x) \) exactly.
In simple words: Factor the divisor first; if all its factors are factors of the polynomial, then the divisor divides the polynomial exactly.

Exam Tip: Breaking quadratic divisors into linear factors makes verification straightforward.

Question 16. Find the values of a and b such that \( f(x) = x^3 + ax^2 + bx + 6 \) when divided by \( (x - 3) \) leaves remainder 3, and has \( (x - 2) \) as a factor.
Answer: From the first condition (remainder is 3 when divided by \( (x - 3) \)):
\( f(3) = 3 \)
\( 27 + 9a + 3b + 6 = 3 \)
\( 9a + 3b = -30 \)
\( 3a + b = -10 \) ... (i)

From the second condition (\( (x - 2) \) is a factor):
\( f(2) = 0 \)
\( 8 + 4a + 2b + 6 = 0 \)
\( 4a + 2b = -14 \)
\( 2a + b = -7 \) ... (ii)

Subtracting (ii) from (i):
\( a = -3 \)

Substituting into (ii):
\( -6 + b = -7 \)
\( b = -1 \)

Therefore, \( a = -3 \) and \( b = -1 \).
In simple words: Combine the Remainder Theorem condition with the Factor Theorem condition to set up two equations.

Exam Tip: Mixed conditions (remainder and factor) can appear together; apply the correct theorem to each.

Exercise 2E

Algebraic Identities (Reference)

Note: The following standard identities are used throughout polynomial factorization:

• \( (a + b)^2 = a^2 + 2ab + b^2 = (a - b)^2 \) is incorrect; actually \( (a - b)^2 = a^2 - 2ab + b^2 \)
• \( (a - b)(a + b) = a^2 - b^2 \)
• \( (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \)
• \( (a + b - c)^2 = a^2 + b^2 + c^2 + 2ab - 2bc - 2ca \)
• \( (a - b + c)^2 = a^2 + b^2 + c^2 - 2ab - 2bc + 2ca \)
• \( (a + b)^3 = a^3 + b^3 + 3ab(a + b) \)
• \( (a - b)^3 = a^3 - b^3 - 3ab(a - b) \)
• \( a^3 + b^3 = (a + b)^3 - 3ab(a + b) = (a + b)(a^2 - ab + b^2) \)
• \( a^3 - b^3 = (a - b)^3 + 3ab(a - b) = (a - b)(a^2 + ab + b^2) \)
• \( a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \)
  If \( a + b + c = 0 \), then \( a^3 + b^3 + c^3 = 3abc \)

Question 1. Factor \( 9x^2 + 12xy \).
Answer: The greatest common factor of 9x² and 12xy is 3x. Factoring out this term:
\( 9x^2 + 12xy = 3x(3x + 4y) \)
In simple words: Find the largest factor that divides every term, then pull it out front.

Exam Tip: Always check for a common monomial factor first—it is the easiest factorization step.

Question 2. Factor \( 18x^2y - 24xyz \).
Answer: The greatest common factor is 6xy. Extracting this factor:
\( 18x^2y - 24xyz = 6xy(3x - 4z) \)
In simple words: Identify all variables and coefficients that divide each term evenly.

Exam Tip: For multi-variable expressions, check each variable separately.

Question 3. Factor \( 27a^3b^3 - 45a^4b^2 \).
Answer: The greatest common factor is 9a³b². Factoring this out:
\( 27a^3b^3 - 45a^4b^2 = 9a^3b^2(3b - 5a) \)
In simple words: Take the smallest exponent for each variable and the greatest common divisor of the coefficients.

Exam Tip: When variables appear in multiple terms, always use the lowest power in the common factor.

Question 4. Factor \( 2a(x + y) - 3b(x + y) \).
Answer: The common factor is \( (x + y) \). Extracting it:
\( 2a(x + y) - 3b(x + y) = (x + y)(2a - 3b) \)
In simple words: When a binomial or larger expression appears in every term, treat it as a single unit and factor it out.

Exam Tip: Grouping by common polynomial factors (not just monomials) is a powerful technique.

Question 5. Factor \( 2x(p^2 + q^2) + 4y(p^2 + q^2) \).
Answer: First extract the monomial common factor 2:
\( 2x(p^2 + q^2) + 4y(p^2 + q^2) = 2[(x)(p^2 + q^2) + 2y(p^2 + q^2)] \)

Now extract the common binomial factor \( (p^2 + q^2) \):
\( = 2(p^2 + q^2)(x + 2y) \)
In simple words: Factor out monomials first, then look for polynomial common factors.

Exam Tip: Always fully simplify by removing all possible common factors at each step.

Question 6. Factor \( x(a - 5) + y(5 - a) \).
Answer: Notice that \( (5 - a) = -(a - 5) \). Rewrite:
\( x(a - 5) + y(5 - a) = x(a - 5) - y(a - 5) = (a - 5)(x - y) \)
In simple words: When factors appear with opposite signs, factor out the negative and flip the sign to reveal the common factor.

Exam Tip: Recognizing opposite binomials as negatives of each other is crucial for factoring by grouping.

Question 7. Factor \( 4(a + b) - 6(a + b)^2 \).
Answer: The common factor is \( (a + b) \). Extract it:
\( 4(a + b) - 6(a + b)^2 = (a + b)[4 - 6(a + b)] = 2(a + b)[2 - 3(a + b)] = 2(a + b)(2 - 3a - 3b) \)
In simple words: When different powers of the same factor appear, use the lowest power in the common factor.

Exam Tip: Simplify by pulling out common numeric factors (like 2) after extracting the polynomial factor.

Question 8. Factor \( 8(3a - 2b)^2 - 10(3a - 2b) \).
Answer: The common factor is \( (3a - 2b) \). Extract it:
\( 8(3a - 2b)^2 - 10(3a - 2b) = (3a - 2b)[8(3a - 2b) - 10] = 2(3a - 2b)[4(3a - 2b) - 5] = 2(3a - 2b)(12a - 8b - 5) \)
In simple words: Pull out the lowest power of the repeated binomial, then simplify.

Exam Tip: Always look for numeric common factors in the remaining expression after factoring the polynomial part.

Question 9. Factor \( x(x + y)^3 - 3x^2y(x + y) \).
Answer: The common factor is \( x(x + y) \). Extract it:
\( x(x + y)^3 - 3x^2y(x + y) = x(x + y)[(x + y)^2 - 3xy] = x(x + y)(x^2 + y^2 + 2xy - 3xy) = x(x + y)(x^2 + y^2 - xy) \)
In simple words: Factor out the common monomials and polynomial factors, then expand and simplify what remains.

Exam Tip: Expand squared terms mentally to combine like terms in the simplified factor.

Question 10. Factor \( x^3 + 2x^2 + 5x + 10 \).
Answer: Group the first two terms and the last two terms:
\( x^3 + 2x^2 + 5x + 10 = x^2(x + 2) + 5(x + 2) = (x + 2)(x^2 + 5) \)
In simple words: Pair terms into groups, factor out the common factor from each pair, then factor the resulting common binomial.

Exam Tip: Factoring by grouping works best when terms share a clear polynomial common factor.

Question 11. Factor \( x^2 + xy - 2xz - 2yz \).
Answer: Group strategically:
\( x^2 + xy - 2xz - 2yz = x(x + y) - 2z(x + y) = (x + y)(x - 2z) \)
In simple words: Arrange and group terms so that each pair has a common factor that leads to a repeated binomial factor.

Exam Tip: When standard grouping (first two, last two) does not work, try alternate pairings.

Question 12. Factor \( a^3b - a^2b + 5ab - 5b \).
Answer: First factor out the common monomial \( b \):
\( a^3b - a^2b + 5ab - 5b = b(a^3 - a^2 + 5a - 5) \)

Now factor by grouping the cubic expression:
\( = b[a^2(a - 1) + 5(a - 1)] = b(a - 1)(a^2 + 5) \)
In simple words: Remove monomial factors first, then apply grouping to what remains.

Exam Tip: Always extract common monomials before attempting polynomial grouping techniques.

Question 13. Factor \( 8 - 4a - 2a^3 + a^4 \).
Answer: Group the terms strategically:
\( 8 - 4a - 2a^3 + a^4 = 4(2 - a) - a^3(2 - a) = (2 - a)(4 - a^3) \)
In simple words: Rearrange if necessary to reveal common factors in grouped pairs.

Exam Tip: Sometimes reordering terms reveals a grouping strategy that is not immediately obvious.

Question 14. Factor \( x^3 - 2x^2y + 3xy^2 - 6y^3 \)
Answer: \( x^2(x - 2y) + 3y^2(x - 2y) = (x - 2y)(x^2 + 3y^2) \)
In simple words: Group terms that share a common factor, extract that factor, and the remaining parts form the second bracket.

Exam Tip: Always look for a common binomial factor after grouping - this signals the factorization is on the right track.

Question 15. Factor \( px + pq - 5q - 5x \)
Answer: \( p(x + q) - 5(q + x) = (x + q)(p - 5) \)
In simple words: Rearrange and group pairs of terms that allow you to pull out common factors, then identify and extract the common binomial.

Exam Tip: Rearranging before grouping can reveal hidden common factors - don't assume the given order is final.

Question 16. Factor \( x^2 - xy + y - x \)
Answer: \( x(x - y) - 1(x - y) = (x - y)(x - 1) \)
In simple words: Factor the first pair and the second pair separately, then pull out the common binomial expression from both.

Exam Tip: When a term appears to stand alone, write it as a coefficient times that term (e.g., \( y = 1 \cdot y \)) to help spot grouping patterns.

Question 17. Factor \( (3a - 1)^2 - 6a + 2 \)
Answer: \( (3a - 1)^2 - 2(3a - 1) = (3a - 1)[(3a - 1) - 2] = (3a - 1)(3a - 3) = 3(3a - 1)(a - 1) \)
In simple words: Treat the squared binomial and the linear term as sharing a common factor, extract it, simplify, and factor further if possible.

Exam Tip: Always check the final result for any remaining common factors - a numerical coefficient may still be factorable.

Question 18. Factor \( (2x - 3)^2 - 8x + 12 \)
Answer: \( (2x - 3)^2 - 4(2x - 3) = (2x - 3)(2x - 3 - 4) = (2x - 3)(2x - 7) \)
In simple words: Identify the common expression in both terms, factor it out, then simplify the remaining bracket.

Exam Tip: Rewrite linear expressions in terms of the binomial present in the squared term to reveal the common factor.

Question 19. Factor \( a^3 + a^2 - 3a^2 - 3 \)
Answer: \( a(a^2 + 1) - 3(a^2 + 1) = (a^2 + 1)(a - 3) \)
In simple words: Group terms to extract a common polynomial, then pull that common polynomial out of both groups.

Exam Tip: Once you identify a common polynomial factor, the second bracket follows directly from the coefficients left behind.

Question 20. Factor \( 3ax - 6ay - 8by + 4bx \)
Answer: \( 3a(x - 2y) + 4b(x - 2y) = (x - 2y)(3a + 4b) \)
In simple words: Rearrange so pairs of terms share a common factor, extract each factor, then identify the common binomial.

Exam Tip: When terms seem unrelated, try rearranging them to create matching binomial structures in each group.

Question 21. Factor \( abx^2 + a^2x + b^2x + ab \)
Answer: \( ax(bx + a) + b(bx + a) = (bx + a)(ax + b) \)
In simple words: Group the first two terms and the last two terms, extract their common factors, then pull out the shared binomial.

Exam Tip: The common binomial in the second step becomes one factor, and the two extracted coefficients form the other factor.

Question 22. Factor \( x^3 - x^2 + ax + x - a - 1 \)
Answer: Rearranging: \( x^3 - x^2 + ax - a + x - 1 \) gives us \( x^2(x - 1) + a(x - 1) + 1(x - 1) = (x - 1)(x^2 + a + 1) \)
In simple words: Organize the terms to reveal a common binomial factor in each group, then extract it to obtain the final factorization.

Exam Tip: Sometimes rearranging multiple times helps - group three or more pairs if needed to find the pattern.

Question 23. Factor \( 2x^3 + 4y^2 - 8xy - 1 \)
Answer: Rearranging: \( 2x - 1 - 8xy + 4y = (2x - 1) - 4y(2x - 1) = (2x - 1)(1 - 4y) \)
In simple words: Reorder terms so a common binomial appears in each grouped pair, then factor it out.

Exam Tip: When exponents vary widely, focus on creating balanced groupings that reveal a repeating binomial.

Question 24. Factor \( ab(x^2 + y^2) - xy(a^2 + b^2) \)
Answer: \( abx^2 + aby^2 - a^2xy - b^2xy \) can be rearranged as \( ax(bx - ay) + by(ay - bx) = ax(bx - ay) - by(bx - ay) = (bx - ay)(ax - by) \)
In simple words: Expand, rearrange to pair terms sharing common binomial factors, then extract those factors systematically.

Exam Tip: Watch for sign changes - sometimes one group needs a negative factor pulled out to match the other group's binomial.

Question 25. Factor \( a^2 + ab(b + 1) + b^3 \)
Answer: \( a^2 + ab^2 + ab + b^3 = a(a + b) + b(a + b) + b^2(a + b) \). Rearranging more carefully: \( a^2 + ab + ab^2 + b^3 = a(a + b) + b^2(a + b) = (a + b)(a + b^2) \)
In simple words: Expand the original expression, regroup to highlight a common binomial, then extract it.

Exam Tip: After expansion, look for terms that naturally pair with each other before assuming a specific grouping.

Question 26. Factor \( a^3 + ab(1 - 2a) - 2b^2 \)
Answer: Expanding: \( a^3 + ab - 2a^2b - 2b^2 = a(a^2 + b) - 2b(a^2 + b) = (a^2 + b)(a - 2b) \)
In simple words: Expand the parentheses, reorder terms to reveal a shared binomial, then pull that binomial out of both groups.

Exam Tip: The common polynomial factor may not be a simple binomial like \( (a + b) \) - it could involve squares or other powers.

Question 27. Factor \( 2a^2 + bc - 2ab - ac \)
Answer: Rearranging: \( 2a^2 - 2ab - ac + bc = 2a(a - b) - c(a - b) = (a - b)(2a - c) \)
In simple words: Reorganize the terms into pairs that each contain a common factor, then extract the shared binomial from both pairs.

Exam Tip: Don't factor the pairs first - group first, then look for the binomial that appears in both groups.

Question 28. Factor \( (ax + by)^2 + (bx - ay)^2 \)
Answer: Expanding: \( a^2x^2 + b^2y^2 + 2abxy + b^2x^2 + a^2y^2 - 2abxy = a^2x^2 + b^2y^2 + b^2x^2 + a^2y^2 = a^2(x^2 + y^2) + b^2(x^2 + y^2) = (a^2 + b^2)(x^2 + y^2) \)
In simple words: Expand both squared terms, combine like terms, then group by the common binomial expression.

Exam Tip: Notice that cross-terms cancel when two squared binomials differ only in the order and signs - this simplifies the factorization.

Question 29. Factor \( a(a + b - c) - bc \)
Answer: \( a^2 + ab - ac - bc = a(a + b) - c(a + b) = (a - c)(a + b) \)
In simple words: Expand the term being multiplied by \( a \), then group pairs of terms that share common factors.

Exam Tip: When a binomial or trinomial is multiplied by a variable, always expand first before attempting to factor.

Question 30. Factor \( a(a - 2b - c) + 2bc \)
Answer: Expanding: \( a^2 - 2ab - ac + 2bc = a(a - 2b) - c(a - 2b) = (a - 2b)(a - c) \)
In simple words: Distribute the outer term, then group and extract common binomial factors from each pair.

Exam Tip: After expansion, the binomial factor that emerges may contain coefficients - be careful to preserve them during grouping.

Question 32. Factor \( ab(x^2 + 1) + x(a^2 + b^2) \)
Answer: Expanding: \( abx^2 + ab + a^2x + b^2x = abx^2 + a^2x + ab + b^2x = ax(bx + a) + b(bx + a) = (bx + a)(ax + b) \)
In simple words: Distribute the products, reorder the resulting terms into two pairs that each contain a common factor, then pull out the shared binomial.

Exam Tip: After expansion, don't stick to the original term order - rearrange freely to create matching patterns in each group.

Question 33. Factor \( x^2 - (a + b)x + ab \)
Answer: \( x^2 - ax - bx + ab = x(x - a) - b(x - a) = (x - a)(x - b) \)
In simple words: Separate the middle term into two parts using the coefficients, group the resulting four terms into pairs, and extract the common binomial.

Exam Tip: For quadratic trinomials, the middle term's coefficient tells you how to split it - look for two numbers that multiply to the constant and add to the middle coefficient.

Exercise 2F

Question 1. Factor \( 25x^2 - 64y^2 \)
Answer: \( (5x)^2 - (8y)^2 = (5x + 8y)(5x - 8y) \). [Since \( a^2 - b^2 = (a + b)(a - b) \)]
In simple words: Recognize the expression as a difference of two perfect squares, then apply the formula to get two linear factors.

Exam Tip: Always check if both terms are perfect squares before trying other factoring methods - this is often the quickest approach.

Question 2. Factor \( 100 - 9x^2 \)
Answer: \( (10)^2 - (3x)^2 = (10 + 3x)(10 - 3x) \). [Since \( a^2 - b^2 = (a + b)(a - b) \)]
In simple words: Write each term as a perfect square, then apply the difference of squares formula to factor.

Exam Tip: Constants like 100 and 4 are perfect squares - recognize them immediately rather than leaving them as numbers.

Question 3. Factor \( 5x^2 - 7y^2 \)
Answer: \( = (\sqrt{5}x)^2 - (\sqrt{7}y)^2 = (\sqrt{5}x + \sqrt{7}y)(\sqrt{5}x - \sqrt{7}y) \). [Since \( a^2 - b^2 = (a + b)(a - b) \)]
In simple words: Even when the coefficients are not perfect squares, you can still express them as squares of square roots and apply the difference of squares formula.

Exam Tip: When coefficients contain square roots or primes, the factors will also contain square roots - this is normal and acceptable.

Question 4. Factor \( (3x + 5y)^2 - 4z^2 \)
Answer: \( (3x + 5y)^2 - (2z)^2 = (3x + 5y + 2z)(3x + 5y - 2z) \). [Since \( a^2 - b^2 = (a + b)(a - b) \)]
In simple words: Treat the entire squared binomial as one "perfect square term" and apply the difference of squares formula.

Exam Tip: The difference of squares formula works even when one or both terms are themselves polynomials or squared expressions.

Question 5. Factor \( 150 - 6x^2 \)
Answer: \( = 6(25 - x^2) = 6(5^2 - x^2) = 6(5 + x)(5 - x) \)
In simple words: First extract any common numerical factor, then apply the difference of squares formula to the remaining binomial.

Exam Tip: Always look for a greatest common factor before using other methods - it simplifies the remaining factorization significantly.

Question 6. Factor \( 20x^2 - 45 \)
Answer: \( = 5(4x^2 - 9) = 5[(2x)^2 - (3)^2] = 5(2x + 3)(2x - 3) \). [Since \( a^2 - b^2 = (a + b)(a - b) \)]
In simple words: Factor out the GCF first, recognize the remaining expression as a difference of squares, and complete the factorization.

Exam Tip: The GCF may include coefficients of the variable terms - don't miss those when factoring them out initially.

Question 7. Factor \( 3x^3 - 48x \)
Answer: \( = 3x(x^2 - 16) = 3x[(x)^2 - (4)^2] = 3x(x + 4)(x - 4) \). [Since \( a^2 - b^2 = (a + b)(a - b) \)]
In simple words: Pull out common factors including variable terms, then apply the difference of squares to what remains.

Exam Tip: When all terms contain a variable, factor it out first - this always reveals additional structure for further factoring.

Question 8. Factor \( 2 - 50x^2 \)
Answer: \( = 2(1 - 25x^2) = 2[(1)^2 - (5x)^2] = 2(1 + 5x)(1 - 5x) \). [Since \( a^2 - b^2 = (a + b)(a - b) \)]
In simple words: Extract the common factor from both terms, then treat the result as a difference of perfect squares.

Exam Tip: The constant 1 is a perfect square equal to \( 1^2 \) - recognize this when it appears in the simplified form.

Question 9. Factor \( 27a^2 - 48b^2 \)
Answer: \( = 3(9a^2 - 16b^2) = 3[(3a)^2 - (4b)^2] = 3(3a + 4b)(3a - 4b) \). [Since \( a^2 - b^2 = (a + b)(a - b) \)]
In simple words: Factor out the common numerical coefficient, recognize the remaining expression as a difference of two squares, and complete the factorization.

Exam Tip: When both terms have the same variable letters raised to the same power, they are both "square-like" and the difference of squares method applies.

Question 10. Factor \( x - 64x^3 \)
Answer: \( = x(1 - 64x^2) = x[(1)^2 - (8x)^2] = x(1 + 8x)(1 - 8x) \). [Since \( a^2 - b^2 = (a + b)(a - b) \)]
In simple words: Extract the common variable factor first, then apply the difference of squares formula to the remaining binomial.

Exam Tip: Even when exponents differ (like \( x \) vs. \( x^3 \)), you can still extract a common factor - pull out the lowest power present in every term.

Question 11. Factor \( 8ab^2 - 18a^3 \)
Answer: \( = 2a(4b^2 - 9a^2) = 2a[(2b)^2 - (3a)^2] = 2a(2b + 3a)(2b - 3a) \). [Since \( a^2 - b^2 = (a + b)(a - b) \)]
In simple words: Remove the common factor from both terms, express what remains as a difference of two perfect squares, and factor completely.

Exam Tip: The GCF includes both numerical and variable parts - for this problem, \( 2a \) is the GCF of \( 8a \) and \( 18a^3 \).

Question 12. Factor \( 6(25 - x^2) \)
Answer: \( = 6(5^2 - x^2) = 6(5 + x)(5 - x) \). [Since \( a^2 - b^2 = (a + b)(a - b) \)]
In simple words: Recognize the expression inside the parentheses as a difference of perfect squares and apply the formula.

Exam Tip: When a difference of squares is already partially grouped, simply apply the factorization formula to that group.

Question 13. Factor \( (a + b)^3 - a^3 - b^3 \)
Answer: Rewriting: \( (a + b)^3 - (a + b) = (a + b)[(a + b)^2 - 1] = (a + b)[(a + b) - 1][(a + b) + 1] = (a + b)(a + b - 1)(a + b + 1) \). [Since \( a^2 - b^2 = (a + b)(a - b) \)]
In simple words: Rewrite to reveal a common factor, extract it, then apply the difference of squares to what remains.

Exam Tip: Cubic and higher-power expressions can still yield to these methods if you first create a difference of perfect squares structure.

Question 14. Factor \( 108a^2 - 3(b - c)^2 \)
Answer: \( = 3[36a^2 - (b - c)^2] = 3[(6a)^2 - (b - c)^2] = 3(6a + b - c)(6a - b + c) \). [Since \( a^2 - b^2 = (a + b)(a - b) \)]
In simple words: Extract the common factor, then treat the squared expressions as "square terms" and apply the difference of squares formula.

Exam Tip: When one term is a squared binomial like \( (b - c)^2 \), treat the entire binomial as a single unit when applying factoring formulas.

Question 15. Factor \( x^3 - 5x^2 - x + 5 \)
Answer: \( = x^2(x - 5) - 1(x - 5) = (x - 5)(x^2 - 1) = (x - 5)(x + 1)(x - 1) \)
In simple words: Group terms to reveal a common binomial factor, extract it, then apply the difference of squares to the remaining polynomial.

Exam Tip: After grouping and extracting a binomial, check if the remaining factor is itself factorable - you may need a second factoring step.

Question 16. Factor \( a^2 + 2ab + b^2 - 9c^2 \)
Answer: \( = (a + b)^2 - (3c)^2 = (a + b + 3c)(a + b - 3c) \). [Since \( a^2 - b^2 = (a + b)(a - b) \)]
In simple words: Recognize the first three terms as a perfect square trinomial, rewrite the expression as a difference of squares, and factor.

Exam Tip: Look for hidden perfect square trinomials - when you see \( a^2 + 2ab + b^2 \), immediately recognize it as \( (a + b)^2 \).

Question 17. Factor \( 9 - a^2 + 2ab - b^2 \)
Answer: \( = 9 - (a^2 - 2ab + b^2) = 3^2 - (a - b)^2 = (3 + a - b)(3 - a + b) \). [Since \( a^2 - b^2 = (a + b)(a - b) \)]
In simple words: Reorganize to reveal a perfect square trinomial, rewrite as a difference of two squares, and apply the formula.

Exam Tip: When terms are in an unusual order, regroup them to recognize perfect square trinomials before attempting other methods.

Question 18. Factor \( a^2 - 4ac + 4c^2 - b^2 \)
Answer: \( = a^2 - 2a(2c) + (2c)^2 - b^2 = (a - 2c)^2 - b^2 = (a - 2c + b)(a - 2c - b) \). [Since \( a^2 - b^2 = (a + b)(a - b) \)]
In simple words: Identify the first three terms as a perfect square trinomial, then treat the result as a difference of two squares.

Exam Tip: Perfect square trinomials have the pattern \( a^2 \pm 2ab + b^2 \) - learn to spot these quickly by checking if the middle term is twice the product of the square roots of the end terms.

Question 19. Factor \( 9a^2 + 3a - 8b - 64b^2 \)
Answer: Rearranging: \( 9a^2 - 64b^2 + 3a - 8b = (3a)^2 - (8b)^2 + (3a - 8b) = (3a + 8b)(3a - 8b) + (3a - 8b) = (3a - 8b)(3a + 8b + 1) \)
In simple words: Organize terms to reveal a difference of two squares plus extra terms, factor the difference of squares, then extract the common binomial from both resulting groups.

Exam Tip: Sometimes you must reorder the terms and use multiple factoring techniques in sequence - difference of squares followed by grouping.

Question 20. Factor \( x^2 - y^2 + 6y - 9 \)
Answer: \( = x^2 - (y^2 - 6y + 9) = x^2 - (y - 3)^2 = [x + (y - 3)][x - (y - 3)] = (x + y - 3)(x - y + 3) \). [Since \( a^2 - b^2 = (a + b)(a - b) \)]
In simple words: Reorganize to identify a perfect square trinomial, then apply the difference of squares formula using the squared trinomial as one term.

Exam Tip: When terms seem scattered, try grouping them mentally to see if hidden perfect squares or perfect square trinomials emerge.

Question 21. Factor \( 4x^2 - 9y^2 - 2x - 3y \)
Answer: Rearranging: \( (2x)^2 - (3y)^2 - (2x + 3y) = (2x + 3y)(2x - 3y) - (2x + 3y) = (2x + 3y)(2x - 3y - 1) \)
In simple words: Reorganize the terms to reveal a difference of squares that shares a binomial factor with a remaining term, then extract the common binomial.

Exam Tip: After factoring a difference of squares, check whether either factor appears elsewhere in the expression - it may be factorable again.

Question 22. Factor \( x^4 - 1 \)
Answer: \( = (x^2)^2 - 1^2 = (x^2 + 1)(x^2 - 1) = (x^2 + 1)(x + 1)(x - 1) \). [Since \( a^2 - b^2 = (a + b)(a - b) \)]
In simple words: Treat the fourth power as a perfect square, apply the difference of squares formula, and check whether the result can be factored further.

Exam Tip: Always continue factoring after the first step - even polynomial factors may be differences of squares themselves, as with \( x^2 - 1 \).

Question 23. Factor \( a - b - a^2 + b^2 \)
Answer: Rearranging: \( (a - b) - (a^2 - b^2) = (a - b) - (a - b)(a + b) = (a - b)[1 - (a + b)] = (a - b)(1 - a - b) \)
In simple words: Reorganize and apply the difference of squares formula to part of the expression, then extract the common binomial factor.

Exam Tip: Sometimes the difference of squares formula is one step in a larger factorization - use it alongside grouping or other techniques.

Question 24. Factor \( x^4 - 625 \)
Answer: \( = (x^2)^2 - (25)^2 = (x^2 + 25)(x^2 - 25) = (x^2 + 25)(x + 5)(x - 5) \). [Since \( a^2 - b^2 = (a + b)(a - b) \)]
In simple words: Rewrite as a difference of two perfect squares, apply the formula, then check if the resulting factors can be factored again.

Exam Tip: The factor \( x^2 + 25 \) cannot be factored further over the real numbers, but \( x^2 - 25 \) is itself a difference of squares requiring additional factorization.

Question 19. Factor \( 9a^2 + 3a - 8b - 64b^2 \)
Answer: Rearranging: \( 9a^2 - 64b^2 + 3a - 8b = (3a)^2 - (8b)^2 + (3a - 8b) = (3a + 8b)(3a - 8b) + (3a - 8b) = (3a - 8b)(3a + 8b + 1) \)
In simple words: Reorder terms to reveal a difference of two squares alongside linear terms, factor the squares, and then extract the shared binomial.

Exam Tip: When a binomial appears as a natural factor after using the difference of squares formula, always check if it also divides the remaining terms.

Question 20. Factor \( x^2 - y^2 + 6y - 9 \)
Answer: \( = x^2 - (y^2 - 6y + 9) = x^2 - (y - 3)^2 = (x + y - 3)(x - y + 3) \)
In simple words: Rearrange the terms to spot a perfect square trinomial, then apply the difference of squares formula with the squared trinomial as one factor.

Exam Tip: Group the terms strategically - it's often worth isolating a perfect square trinomial from the rest of the expression.

Question 21. Factor \( 4x^2 - 9y^2 - 2x - 3y \)
Answer: Rearranging: \( (2x + 3y)(2x - 3y) - (2x + 3y) = (2x + 3y)(2x - 3y - 1) \)
In simple words: Reorganize to create a difference of two squares and another term, apply the difference of squares formula, then factor out the common binomial.

Exam Tip: After factoring a difference of squares, always scan the remaining expression for factors that might be present elsewhere - this reveals a second level of factorization.

Question 22. Factor \( x^4 - 1 \)
Answer: \( = (x^2 + 1)(x^2 - 1) = (x^2 + 1)(x + 1)(x - 1) \)
In simple words: Apply the difference of squares formula twice - first to the fourth power expression, then to the resulting quadratic that is also a difference of squares.

Exam Tip: Higher even powers can always be written as perfect squares - treat \( x^4 \) as \( (x^2)^2 \) and \( x^6 \) as \( (x^3)^2 \), for example.

Question 23. Factor \( a - b - a^2 + b^2 \)
Answer: Rearranging: \( (a - b) - (a^2 - b^2) = (a - b) - (a - b)(a + b) = (a - b)(1 - a - b) \)
In simple words: Separate the expression into parts, apply the difference of squares formula to one part, then identify and extract the common factor.

Exam Tip: When one factor appears in multiple parts of an expression, factor it out globally - this simplifies the final answer.

Question 24. Factor \( x^4 - 625 \)
Answer: \( = (x^2 + 25)(x^2 - 25) = (x^2 + 25)(x + 5)(x - 5) \)
In simple words: Write as a difference of two perfect squares, apply the formula, and then check if any resulting factors can be factored further using the same method.

Exam Tip: The sum of squares \( x^2 + 25 \) does not factor over the reals, but the difference \( x^2 - 25 \) must always be factored - never leave it as a single factor.

Exercise 2G

Question 1. Factor \( x^2 + 11x + 30 \)
Answer: \( = x^2 + 6x + 5x + 30 = x(x + 6) + 5(x + 6) = (x + 6)(x + 5) \)
In simple words: Split the middle term into two parts whose coefficients add up to 11 and multiply to give 30, group into pairs, and extract common factors.

Exam Tip: For a quadratic \( x^2 + bx + c \), find two numbers that multiply to \( c \) and add to \( b \) - then split the middle term using those two numbers.

Question 2. Factor \( x^2 + 18x + 32 \)
Answer: \( = x^2 + 16x + 2x + 32 = x(x + 16) + 2(x + 16) = (x + 16)(x + 2) \)
In simple words: Identify two numbers that multiply to 32 and add to 18 - they are 16 and 2 - use these to split the middle term and factor by grouping.

Exam Tip: Always verify your two numbers: do they multiply to the constant and add to the middle coefficient? If yes, the factorization will succeed.

Question 3. Factor \( x^2 + 7x - 18 \)
Answer: \( = x^2 + 9x - 2x - 18 = x(x + 9) - 2(x + 9) = (x + 9)(x - 2) \)
In simple words: Find two numbers that multiply to -18 and add to 7 - they are 9 and -2 - split the middle term and factor by grouping.

Exam Tip: With a negative constant, one of the two numbers must be negative - their product is negative while their sum matches the middle coefficient.

Question 4. Factor \( x^2 + 5x - 6 \)
Answer: \( = x^2 + 6x - x - 6 = x(x + 6) - 1(x + 6) = (x + 6)(x - 1) \)
In simple words: Locate two numbers multiplying to -6 and summing to 5 - they are 6 and -1 - split the middle term accordingly and factor.

Exam Tip: Writing a term like \( -x \) explicitly as \( -1(x) \) helps you see the common binomial factor clearly.

Question 5. Factor \( y^2 - 4y + 3 \)
Answer: \( = y^2 - 3y - y + 3 = y(y - 3) - 1(y - 3) = (y - 3)(y - 1) \)
In simple words: Find two numbers that multiply to 3 and add to -4 - they are -3 and -1 - split the middle term and group to factor.

Exam Tip: When both the constant and middle coefficient are negative, both numbers in your pair must be negative.

Question 6. Factor \( x^2 - 21x + 108 \)
Answer: \( = x^2 - 12x - 9x + 108 = x(x - 12) - 9(x - 12) = (x - 12)(x - 9) \)
In simple words: Identify two numbers multiplying to 108 and adding to -21 - they are -12 and -9 - split and factor by grouping.

Exam Tip: When factoring 108, consider factor pairs like 1×108, 2×54, 3×36, 4×27, 6×18, 9×12 - one pair will sum to -21.

Question 7. Factor \( x^2 - 11x - 80 \)
Answer: \( = x^2 - 16x + 5x - 80 = x(x - 16) + 5(x - 16) = (x - 16)(x + 5) \)
In simple words: Find two numbers that multiply to -80 and add to -11 - they are -16 and 5 - use these to split the middle term and complete the factorization.

Exam Tip: Negative factor pairs are often needed - if one number is much larger in absolute value than the other, their difference will match the middle coefficient.

Question 8. Factor \( x^2 - x - 156 \)
Answer: \( = x^2 - 13x + 12x - 156 = x(x - 13) + 12(x - 13) = (x - 13)(x + 12) \)
In simple words: Locate two numbers that multiply to -156 and sum to -1 - they are -13 and 12 - split the middle term and factor.

Exam Tip: For larger constants, listing factor pairs systematically helps - multiply each potential divisor by 156 to check if two factors sum to the middle coefficient.

Question 9. Factor \( z^2 - 32z - 105 \)
Answer: \( = z^2 - 35z + 3z - 105 = z(z - 35) + 3(z - 35) = (z - 35)(z + 3) \)
In simple words: Find two numbers that multiply to -105 and add to -32 - they are -35 and 3 - split the middle term and extract the common binomial.

Exam Tip: Verify your factorization by expanding: \( (z - 35)(z + 3) = z^2 + 3z - 35z - 105 = z^2 - 32z - 105 \) ✓

Question 10. Factor \( 40 + 3x - x^2 \)
Answer: Rearranging: \( -x^2 + 3x + 40 = -(x^2 - 3x - 40) = -(x^2 - 8x + 5x - 40) \) or directly: \( = 40 + 8x - 5x - x^2 = 8(5 + x) - x(5 + x) = (5 + x)(8 - x) \)
In simple words: Reorder the terms in standard form, identify two numbers that multiply to -40 and add to 3 - they are 8 and -5 - split and factor by grouping.

Exam Tip: When the leading coefficient is negative, either factor out -1 first or rearrange the original terms to place the quadratic in standard form before factoring.

Question 11. Factor \( 6 - x - x^2 \)
Answer: Rearranging: \( -x^2 - x + 6 = -(x^2 + x - 6) = -(x^2 + 3x - 2x - 6) \). Alternatively, directly: \( = 6 + 2x - 3x - x^2 = 2(3 + x) - x(3 + x) = (3 + x)(2 - x) \)
In simple words: Rewrite in standard form, find two numbers that multiply to -6 and add to -1 - they are 2 and -3 - group the terms accordingly and extract common factors.

Exam Tip: For expressions like \( a + bx + cx^2 \), rearrange to standard form \( cx^2 + bx + a \) before applying the splitting method.

Question 12. Factor \( 7x^2 + 49x + 84 \)
Answer: \( = 7(x^2 + 7x + 12) = 7[x^2 + 4x + 3x + 12] = 7[x(x + 4) + 3(x + 4)] = 7(x + 4)(x + 3) \)
In simple words: First factor out the GCF of 7, then split the middle term of the remaining quadratic using numbers that multiply to 12 and add to 7 - these are 4 and 3.

Exam Tip: Always look for a GCF in all terms before using the splitting method - removing it simplifies the subsequent factorization significantly.

Question 13. Factor \( m^2 + 17mn - 84n^2 \)
Answer: \( = m^2 + 21mn - 4mn - 84n^2 = m(m + 21n) - 4n(m + 21n) = (m + 21n)(m - 4n) \)
In simple words: Find two numbers that multiply to -84 and add to 17 - they are 21 and -4 - split the middle term and group to extract the common binomial factor.

Exam Tip: When two variables are present, the constant term is the product of their squared coefficients - here, the constant is \( -84n^2 \), so you multiply the squared part by the factorization of -84.

Question 14. Factor \( 5x^2 + 16x + 3 \)
Answer: \( = 5x^2 + 15x + x + 3 = 5x(x + 3) + 1(x + 3) = (5x + 1)(x + 3) \)
In simple words: Find two numbers that multiply to \( 5 \times 3 = 15 \) and add to 16 - they are 15 and 1 - split the middle term and factor by grouping.

Exam Tip: When the leading coefficient is not 1, multiply it by the constant to find the target product for splitting - here, \( 5 \times 3 = 15 \).

Question 15. Factor \( 6x^2 + 17x + 12 \)
Answer: \( = 6x^2 + 9x + 8x + 12 = 3x(2x + 3) + 4(2x + 3) = (2x + 3)(3x + 4) \)
In simple words: Find two numbers that multiply to \( 6 \times 12 = 72 \) and add to 17 - they are 9 and 8 - split and factor by grouping.

Exam Tip: For a trinomial \( ax^2 + bx + c \), split using numbers whose product is \( ac \) - this approach works regardless of the value of \( a \).

Question 16. Factor \( 9x^2 + 18x + 8 \)
Answer: \( = 9x^2 + 12x + 6x + 8 = 3x(3x + 4) + 2(3x + 4) = (3x + 4)(3x + 2) \)
In simple words: Find two numbers that multiply to \( 9 \times 8 = 72 \) and add to 18 - they are 12 and 6 - split the middle term and extract the common binomial.

Exam Tip: When both splits produce the same binomial factor in different groups, you've correctly identified the factorization.

Question 17. Factor \( 14x^2 + 9x + 1 \)
Answer: \( = 14x^2 + 7x + 2x + 1 = 7x(2x + 1) + (2x + 1) = (7x + 1)(2x + 1) \)
In simple words: Identify two numbers that multiply to \( 14 \times 1 = 14 \) and add to 9 - they are 7 and 2 - split and group to obtain the final factorization.

Exam Tip: When a term appears to stand alone after factoring one pair (like the lone "(2x + 1)" above), its coefficient is 1 - always make this explicit.

Question 18. Factor \( 2x^2 + 3x - 90 \)
Answer: \( = 2x^2 - 12x + 15x - 90 = 2x(x - 6) + 15(x - 6) = (x - 6)(2x + 15) \)
In simple words: Find two numbers that multiply to \( 2 \times (-90) = -180 \) and add to 3 - they are -12 and 15 - split and factor by grouping.

Exam Tip: With a negative constant, one number in the pair is negative - choose the pair whose difference equals the middle coefficient when considering their sum.

Question 19. Factor \( 2x^2 + 11x - 21 \)
Answer: \( = 2x^2 + 14x - 3x - 21 = 2x(x + 7) - 3(x + 7) = (x + 7)(2x - 3) \)
In simple words: Find two numbers that multiply to \( 2 \times (-21) = -42 \) and add to 11 - they are 14 and -3 - split and extract the common binomial.

Exam Tip: After factoring by grouping, verify that the two binomial factors are distinct - if they're identical, you may have made an arithmetic error.

Question 20. Factor \( 3x^2 - 14x + 8 \)
Answer: \( = 3x^2 - 12x - 2x + 8 = 3x(x - 4) - 2(x - 4) = (x - 4)(3x - 2) \)
In simple words: Find two numbers that multiply to \( 3 \times 8 = 24 \) and add to -14 - they are -12 and -2 - split and factor by grouping.

Exam Tip: When both coefficients in the middle-term split are negative, the trinomial will have two negative factors or a different configuration - check your sum carefully.

Question 21. Factor \( 18x^2 + 3x - 10 \)
Answer: \( = 18x^2 - 12x + 15x - 10 = 6x(3x - 2) + 5(3x - 2) = (6x + 5)(3x - 2) \)
In simple words: Find two numbers that multiply to \( 18 \times (-10) = -180 \) and add to 3 - they are -12 and 15 - split the middle term and group to extract the common binomial.

Exam Tip: The GCF of the coefficients in each grouped pair reveals the factors - here, 6 and 5 become part of the final binomial factors.

Question 22. Factor \( 15x^2 + 2x - 8 \)
Answer: \( = 15x^2 - 10x + 12x - 8 = 5x(3x - 2) + 4(3x - 2) = (3x - 2)(5x + 4) \)
In simple words: Find two numbers that multiply to \( 15 \times (-8) = -120 \) and add to 2 - they are -10 and 12 - split and group to obtain the factorization.

Exam Tip: Organize your factor pairs systematically when seeking two numbers with a specific product and sum - this saves time and reduces errors.

Question 23. Factor \( 6x^2 + 11x - 10 \)
Answer: \( = 6x^2 + 15x - 4x - 10 = 3x(2x + 5) - 2(2x + 5) = (2x + 5)(3x - 2) \)
In simple words: Identify two numbers that multiply to \( 6 \times (-10) = -60 \) and add to 11 - they are 15 and -4 - split the middle term and extract the common binomial factor.

Exam Tip: Cross-check your factorization using FOIL: \( (2x + 5)(3x - 2) = 6x^2 - 4x + 15x - 10 = 6x^2 + 11x - 10 \) ✓

Question 24. Factor \( 30x^2 + 7x - 15 \)
Answer: \( = 30x^2 - 18x + 25x - 15 = 6x(5x - 3) + 5(5x - 3) = (5x - 3)(6x + 5) \)
In simple words: Find two numbers that multiply to \( 30 \times (-15) = -450 \) and add to 7 - they are -18 and 25 - split and factor by grouping to complete the factorization.

Exam Tip: For larger products like -450, break it into prime factors to find all factor pairs more efficiently.

Question 25. Factor \( 24x^2 - 41x + 12 \)
Answer: \( = 24x^2 - 32x - 9x + 12 = 8x(3x - 4) - 3(3x - 4) = (3x - 4)(8x - 3) \)
In simple words: Find two numbers that multiply to \( 24 \times 12 = 288 \) and add to -41 - they are -32 and -9 - split the middle term and extract the common binomial.

Exam Tip: When both coefficients are negative and their product is positive, both numbers in the pair must be negative.

Question 26. Factor \( 2x^2 - 7x - 15 \)
Answer: \( = 2x^2 - 10x + 3x - 15 = 2x(x - 5) + 3(x - 5) = (x - 5)(2x + 3) \)
In simple words: Find two numbers that multiply to \( 2 \times (-15) = -30 \) and add to -7 - they are -10 and 3 - split and group to obtain the final factorization.

Exam Tip: Always verify by expanding your factors back to the original trinomial - this is a quick check for correctness.

Question 27. Factor \( 6x^2 - 5x - 21 \)
Answer: \( = 6x^2 + 9x - 14x - 21 = 3x(2x + 3) - 7(2x + 3) = (3x - 7)(2x + 3) \)
In simple words: Identify two numbers that multiply to \( 6 \times (-21) = -126 \) and add to -5 - they are 9 and -14 - split the middle term and factor by grouping.

Exam Tip: When the GCF of each grouped pair yields coefficients that appear in the other group's binomial, your factorization is correct.

Question 28. Factor \( 10x^2 - 9x - 7 \)
Answer: \( = 10x^2 + 5x - 14x - 7 = 5x(2x + 1) - 7(2x + 1) = (2x + 1)(5x - 7) \)
In simple words: Find two numbers that multiply to \( 10 \times (-7) = -70 \) and add to -9 - they are 5 and -14 - split and extract the common binomial factor.

Exam Tip: Negative factors appear when the constant is negative - organize the pairs so the larger absolute value is paired with the negative sign.

Question 29. Factor \( 5x^2 - 16x - 21 \)
Answer: \( = 5x^2 + 5x - 21x - 21 = 5x(x + 1) - 21(x + 1) = (x + 1)(5x - 21) \)
In simple words: Find two numbers that multiply to \( 5 \times (-21) = -105 \) and add to -16 - they are 5 and -21 - split the middle term and group to complete the factorization.

Exam Tip: When one factor in a pair is the entire polynomial's constant term, you've likely found the correct split.

Question 30. Factor \( 2x^2 - x - 21 \)
Answer: \( = 2x^2 + 6x - 7x - 21 = 2x(x + 3) - 7(x + 3) = (x + 3)(2x - 7) \)
In simple words: Find two numbers that multiply to \( 2 \times (-21) = -42 \) and add to -1 - they are 6 and -7 - split and factor by grouping.

Exam Tip: A middle coefficient of -1 means the two numbers differ by 1 in absolute value - one is one larger than the other.

Question 31. Factor \( 15x^2 - x - 28 \)
Answer: \( = 15x^2 + 20x - 21x - 28 = 5x(3x + 4) - 7(3x + 4) = (3x + 4)(5x - 7) \)
In simple words: Find two numbers that multiply to \( 15 \times (-28) = -420 \) and add to -1 - they are 20 and -21 - split and extract the common binomial.

Exam Tip: Larger products like -420 may require listing prime factorizations first to find factor pairs efficiently.

Question 32. Factor \( 8a^2 - 27ab + 9b^2 \)
Answer: \( = 8a^2 - 24ab - 3ab + 9b^2 = 8a(a - 3b) - 3b(a - 3b) = (a - 3b)(8a - 3b) \)
In simple words: Find two numbers that multiply to \( 8 \times 9 = 72 \) and add to -27 - they are -24 and -3 - split the middle term and group to obtain the factorization.

Exam Tip: When both variables are present, the same splitting method applies - the binomial factors will contain both variables as well.

Question 33. Factor \( 5x^2 + 33xy - 14y^2 \)
Answer: \( = 5x^2 + 35xy - 2xy - 14y^2 = 5x(x + 7y) - 2y(x + 7y) = (x + 7y)(5x - 2y) \)
In simple words: Find two numbers that multiply to \( 5 \times (-14) = -70 \) and add to 33 - they are 35 and -2 - split the middle term and extract the common binomial.

Exam Tip: Trinomials with two variables factor identically to those with one variable - the process and logic are unchanged.

Question 34. Factor \( 3x^3 - x^2 - 10x \)
Answer: \( = x(3x^2 - x - 10) = x[3x^2 - 6x + 5x - 10] = x[3x(x - 2) + 5(x - 2)] = x(x - 2)(3x + 5) \)
In simple words: First factor out the common factor \( x \), then factor the remaining quadratic by splitting the middle term using numbers that multiply to -30 and add to -1 - these are -6 and 5.

Exam Tip: Always extract common factors from all terms first - this simplifies the remaining factorization and reveals the complete factorization.

Question 35. Factor \( \frac{1}{3}x^2 - 2x - 9 \)
Answer: \( = \frac{1}{3}(x^2 - 6x - 27) = \frac{1}{3}(x^2 - 9x + 3x - 27) = \frac{1}{3}[x(x - 9) + 3(x - 9)] = \frac{1}{3}(x - 9)(x + 3) \)
In simple words: Factor out the coefficient of \( x^2 \), then split the resulting quadratic using numbers that multiply to -27 and add to -6 - these are -9 and 3.

Exam Tip: Fractional coefficients can be handled by factoring them out as a GCF - this leaves an integer quadratic that is easier to work with.

Question 36. Factor \( x^2 - 2x + \frac{7}{16} \)
Answer: \( = \frac{1}{16}(16x^2 - 32x + 7) = \frac{1}{16}(16x^2 - 4x - 28x + 7) = \frac{1}{16}[4x(4x - 1) - 7(4x - 1)] = \frac{1}{16}(4x - 1)(4x - 7) \)
In simple words: Multiply through by 16 to clear the fraction, factor the resulting quadratic by splitting using numbers that multiply to 112 and add to -32 - these are -4 and -28 - then divide back by 16.

Exam Tip: Clearing fractions first simplifies the factorization - remember to incorporate the reciprocal of your clearing factor into the final answer.

Question 37. Factor \( \sqrt{2}x^2 + 3x + \sqrt{2} \)
Answer: \( = \sqrt{2}x^2 + 2x + x + \sqrt{2} = \sqrt{2}x(x + \sqrt{2}) + 1(x + \sqrt{2}) = (x + \sqrt{2})(\sqrt{2}x + 1) \)
In simple words: Split the middle term using numbers that multiply to \( \sqrt{2} \times \sqrt{2} = 2 \) and add to 3 - these are 2 and 1 - then group and extract the common binomial.

Exam Tip: Irrational coefficients follow the same factorization rules - the splitting method works identically whether coefficients are rational or irrational.

Question 38. Factor \( \sqrt{5}x^2 + 2x - 3\sqrt{5} \)
Answer: \( = \sqrt{5}x^2 + 5x - 3x - 3\sqrt{5} = \sqrt{5}x(x + \sqrt{5}) - 3(x + \sqrt{5}) = (x + \sqrt{5})(\sqrt{5}x - 3) \)
In simple words: Find two numbers that multiply to \( \sqrt{5} \times (-3\sqrt{5}) = -15 \) and add to 2 - they are 5 and -3 - split the middle term and group to complete the factorization.

Exam Tip: When the constant contains a square root, multiply the leading coefficient by the coefficient of the square root (ignoring the radical) to find the product target for splitting.

Question 39. Factor \( 2x^2 + 3\sqrt{3}x + 3 \)
Answer: \( = 2x^2 + 2\sqrt{3}x + \sqrt{3}x + 3 = 2x(x + \sqrt{3}) + \sqrt{3}(x + \sqrt{3}) = (x + \sqrt{3})(2x + \sqrt{3}) \)
In simple words: Split the middle term using numbers that multiply to 6 and add to \( 3\sqrt{3} \) - they are \( 2\sqrt{3} \) and \( \sqrt{3} \) - group and extract the common binomial factor.

Exam Tip: When coefficients contain surds, express them in terms of the surd for easier recognition of factors - here, \( 2\sqrt{3} + \sqrt{3} = 3\sqrt{3} \) ✓

Question 40. Factor \( 2\sqrt{3}x^2 + x - 5\sqrt{3} \)
Answer: \( = 2\sqrt{3}x^2 + 6x - 5x - 5\sqrt{3} = 2\sqrt{3}x(x + \sqrt{3}) - 5(x + \sqrt{3}) = (x + \sqrt{3})(2\sqrt{3}x - 5) \)
In simple words: Find two numbers that multiply to \( 2\sqrt{3} \times (-5\sqrt{3}) = -30 \) and add to 1 - they are 6 and -5 - split the middle term and group to obtain the factorization.

Exam Tip: Multiplying a coefficient with a surd by a constant gives: \( 2\sqrt{3} \times (-5\sqrt{3}) = -10 \times 3 = -30 \), which is a rational target for finding the splitting numbers.

Question 41. Factor \( 5\sqrt{5}x^2 + 20x + 3\sqrt{5} \)
Answer: \( = 5\sqrt{5}x^2 + 15x + 5x + 3\sqrt{5} = 5\sqrt{5}x(x + \sqrt{5}) + \sqrt{5}(x + \sqrt{5}) = (x + \sqrt{5})(5\sqrt{5}x + \sqrt{5}) \)
In simple words: Find two numbers that multiply to \( 5\sqrt{5} \times 3\sqrt{5} = 75 \) and add to 20 - they are 15 and 5 - split the middle term and extract the common binomial factor.

Exam Tip: The product \( 5\sqrt{5} \times 3\sqrt{5} = 5 \times 3 \times 5 = 75 \) (the radical parts multiply to give 5) - use this to find rational splitting numbers.

Question 42. Factor \( 7\sqrt{2}x^2 - 10x - 4\sqrt{2} \)
Answer: \( = 7\sqrt{2}x^2 - 14x + 4x - 4\sqrt{2} = 7\sqrt{2}x(x - \sqrt{2}) + 4(x - \sqrt{2}) = (x - \sqrt{2})(7\sqrt{2}x + 4) \)
In simple words: Find two numbers that multiply to \( 7\sqrt{2} \times (-4\sqrt{2}) = -56 \) and add to -10 - they are -14 and 4 - split the middle term and group to complete the factorization.

Exam Tip: After grouping, verify that each group yields the same binomial factor - this confirms the factorization is correct.

Question 43. Factor \( 6\sqrt{3}x^2 - 47x + 5\sqrt{3} \)
Answer: \( = 6\sqrt{3}x^2 - 45x - 2x + 5\sqrt{3} = 3\sqrt{3}x(2x - 5\sqrt{3}) - 1(2x - 5\sqrt{3}) = (2x - 5\sqrt{3})(3\sqrt{3}x - 1) \)
In simple words: Find two numbers that multiply to \( 6\sqrt{3} \times 5\sqrt{3} = 90 \) and add to -47 - they are -45 and -2 - split the middle term and group to obtain the final answer.

Exam Tip: Always double-check your arithmetic when dealing with surds - a small error in calculating the product or sum can lead to an incorrect splitting.

Question 44. Factor \( 7x^2 + 2\sqrt{14}x + 2 \)
Answer: \( = 7x^2 + \sqrt{14}x + \sqrt{14}x + 2 = \sqrt{7}x(\sqrt{7}x + \sqrt{2}) + \sqrt{2}(\sqrt{7}x + \sqrt{2}) = (\sqrt{7}x + \sqrt{2})(\sqrt{7}x + \sqrt{2}) \)
In simple words: The expression factors as a perfect square - split using numbers that multiply to 14 and add to \( 2\sqrt{14} \) - these are \( \sqrt{14} \) and \( \sqrt{14} \) - which simplifies to \( (\sqrt{7}x + \sqrt{2})^2 \).

Exam Tip: When the two splitting numbers are identical, the trinomial is a perfect square - the factorization is of the form \( (ax + b)^2 \).

Question 45. Factor \( 2(x + y)^2 - 9(x + y) - 5 \)
Answer: Let \( z = x + y \). Then, \( 2z^2 - 9z - 5 = 2z^2 - 10z + z - 5 = 2z(z - 5) + 1(z - 5) = (z - 5)(2z + 1) \). Now, replacing \( z \) by \( (x + y) \), we get: \( [x + y - 5][2(x + y) + 1] = (x + y - 5)(2x + 2y + 1) \)
In simple words: Substitute the binomial with a single variable to simplify, factor the resulting quadratic, then substitute back to obtain the final factorization in terms of the original variables.

Exam Tip: Substitution transforms complex-looking expressions into standard quadratics - always remember to substitute back at the end.

Question 46. Factor \( 9(2a - b)^2 - 4(2a - b) - 13 \)
Answer: Let \( c = 2a - b \). Then, \( 9c^2 - 4c - 13 = 9c^2 - 13c + 9c - 13 = c(9c - 13) + 1(9c - 13) = (c + 1)(9c - 13) \). Now, replacing \( c \) by \( (2a - b) \), we get: \( 9(2a - b)^2 - 4(2a - b) - 13 = (2a - b + 1)[9(2a - b) - 13] = (2a - b + 1)(18a - 9b - 13) \)
In simple words: Set the repeated expression equal to a new variable, factor the resulting simpler polynomial, and then replace the variable back with the original binomial.

Exam Tip: This substitution method is powerful for expressions where a binomial or trinomial is repeated - it always simplifies the factorization significantly.

Question 47. Factor \( 7(x - 2y)^2 - 25(x - 2y) + 12 \)
Answer: Let \( z = x - 2y \). Then, \( 7z^2 - 25z + 12 = 7z^2 - 21z - 4z + 12 = 7z(z - 3) - 4(z - 3) = (z - 3)(7z - 4) \). Now, replacing \( z \) by \( (x - 2y) \), we get: \( 7(x - 2y)^2 - 25(x - 2y) + 12 = (x - 2y - 3)[7(x - 2y) - 4] = (x - 2y - 3)(7x - 14y - 4) \)
In simple words: Substitute the common binomial expression with a single letter, factor the resulting quadratic, and then revert to the original variables.

Exam Tip: Substitution is particularly useful when the same complex binomial appears as the argument of a polynomial - it removes the complexity and reveals the underlying factorization structure.

Question 48. Factor \( 4x^4 + 7x^2 - 2 \)
Answer: Let \( y = x^2 \). Then, \( 4y^2 + 7y - 2 = 4y^2 + 8y - y - 2 = 4y(y + 2) - 1(y + 2) = (y + 2)(4y - 1) \). Now, replacing \( y \) by \( x^2 \), we get: \( (x^2 + 2)(4x^2 - 1) = (x^2 + 2)(2x + 1)(2x - 1) \)
In simple words: Substitute the even power of the variable with a new variable to create a standard quadratic, factor it, substitute back, and check if any resulting factors can be factored further.

Exam Tip: Expressions in terms of even powers (like \( x^4 \) or \( x^2 \)) often benefit from substitution - set the lower even power equal to a new variable to transform into a quadratic.

Exercise 2H

Formula Summary:

Question 1. Expand (a + 2b + 5c)², (2a - b + c)², and (a - 2b - 3c)²
Answer:
(i) \( (a + 2b + 5c)^2 = a^2 + 4b^2 + 25c^2 + 4ab + 20bc + 10ac \)
(ii) \( (2a - b + c)^2 = 4a^2 + b^2 + c^2 - 4ab - 2bc + 4ac \)
(iii) \( (a - 2b - 3c)^2 = a^2 + 4b^2 + 9c^2 - 4ab + 12bc - 6ac \)
In simple words: Apply the formula \( (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \), carefully tracking the signs of each term and the cross products.

Exam Tip: Organize your working in the same order: squares first, then cross products - this reduces arithmetic errors and makes verification easier.

Question 2. Expand (2a - 5b - 7c)², (-3a + 4b - 5c)², and use the formula to verify.
Answer:
(i) \( (2a - 5b - 7c)^2 = 4a^2 + 25b^2 + 49c^2 - 20ab + 70bc - 28ac \)
(ii) \( (-3a + 4b - 5c)^2 = 9a^2 + 16b^2 + 25c^2 - 24ab - 40bc + 30ac \)
In simple words: For each trinomial, square each term, then add twice the product of each pair of terms, watching the signs carefully based on the original expression.

Exam Tip: When all terms in a trinomial are negative or have mixed signs, the cross product terms will have specific sign patterns - work slowly through each one.

Question 3. Express 4x² + 9y² + 16z² + 12xy - 24yz - 16xz as a perfect square trinomial.
Answer: \( = (2x)^2 + (3y)^2 + (-4z)^2 + 2(2x)(3y) + 2(3y)(-4z) + 2(-4z)(2x) = (2x + 3y - 4z)^2 \)
In simple words: Identify the square roots of the squared terms, verify that the cross products match the given middle terms, and write the result as a perfect square.

Exam Tip: Always verify by expanding your perfect square back to the original expression - this confirms your identification is correct.

Question 4. Express 9x² + 16y² + 4z² - 24xy + 16yz - 12xz as a perfect square trinomial.
Answer: \( = (-3x)^2 + (4y)^2 + (2z)^2 + 2(-3x)(4y) + 2(4y)(2z) + 2(2z)(-3x) = (-3x + 4y + 2z)^2 \)
In simple words: Find the square roots of each squared term, calculate the required cross products, and confirm they match the given middle terms before writing the perfect square form.

Exam Tip: The order of the terms inside the square does not matter - \( (-3x + 4y + 2z)^2 \) equals \( (4y + 2z - 3x)^2 \), etc.

Question 5. Express 25x² + 4y² + 9z² - 20xy - 12yz + 30xz as a perfect square trinomial.
Answer: \( = (5x)^2 + (-2y)^2 + (3z)^2 + 2(5x)(-2y) + 2(-2y)(3z) + 2(3z)(5x) = (5x - 2y + 3z)^2 \)
In simple words: Extract the square roots of each term, construct the cross products, and verify they match the given expression before finalizing the perfect square form.

Exam Tip: If any cross product doesn't match, you've chosen the wrong sign for one of the square roots - recalculate with opposite signs.

Question 6. Evaluate (99)² and (998)² using algebraic formulas.
Answer:
(i) \( (99)^2 = (100 - 1)^2 = 10000 - 200 + 1 = 9801 \)
(ii) \( (998)^2 = (1000 - 2)^2 = 1000000 - 4000 + 4 = 996004 \)
In simple words: Express each number as a binomial close to a perfect power, apply the formula \( (a - b)^2 = a^2 - 2ab + b^2 \), and evaluate.

Exam Tip: Choosing a nearby perfect power (like 100 for 99, or 1000 for 998) makes the calculation much quicker than direct multiplication.

Exercise 2I

Question 1. Expand the following and verify the result for the given values.
(i) \( (2a - 5b - 7c)^2 \)
(ii) \( (-3a + 4b - 5c)^2 \)
(iii) \( \left(\frac{1}{2}a - \frac{1}{4}b + 2\right)^2 \)

Answer:
(i) \( (2a - 5b - 7c)^2 = 4a^2 + 25b^2 + 49c^2 - 20ab + 70bc - 28ac \)
(ii) \( (-3a + 4b - 5c)^2 = 9a^2 + 16b^2 + 25c^2 - 24ab - 40bc + 30ac \)
(iii) \( \left(\frac{1}{2}a - \frac{1}{4}b + 2\right)^2 = \frac{a^2}{4} + \frac{b^2}{16} + 4 - \frac{ab}{4} - b + 2a \)
In simple words: Apply the perfect square trinomial formula to expand each expression, organizing the results as squares of individual terms followed by cross products with proper signs.

Exam Tip: For fractional coefficients, expand carefully - each squared term and cross product must be computed separately to avoid combining unlike terms.

Question 2. Express each expression as a perfect square and verify.
Answer: Follow the method shown in Exercise 2H - identify the square roots of squared terms, calculate cross products, and confirm they match the given expression before writing the perfect square form.
In simple words: Work backwards from the expanded form - identify the three base terms by taking square roots, then verify all cross products align with the given coefficients.

Exam Tip: Always expand your proposed perfect square to double-check against the original expression - a single sign error invalidates the entire answer.

Question 3. Expand 4x² + 9y² + 16z² + 12xy - 24yz - 16xz and express as a perfect square.
Answer: \( (2x + 3y - 4z)^2 \)
In simple words: Recognize the pattern of a perfect square trinomial - find the three base terms by identifying the square roots of the coefficient squares, verify the cross products, and write the final form.

Exam Tip: When verifying, expand your answer step-by-step: \( (2x + 3y - 4z)^2 = (2x)^2 + (3y)^2 + (-4z)^2 + 2(2x)(3y) + 2(3y)(-4z) + 2(2x)(-4z) \) to match the original.

Question 4. Expand 9x² + 16y² + 4z² - 24xy + 16yz - 12xz and express as a perfect square.
Answer: \( (-3x + 4y + 2z)^2 \)
In simple words: Work through the square root identifications, compute the expected cross products, match them against the given expression, and finalize the perfect square form.

Exam Tip: Notice the negative sign in the first term - when expanding \( (-3x + 4y + 2z)^2 \), the negative carries through into the cross products involving \( -3x \).

Question 5. Expand 25x² + 4y² + 9z² - 20xy - 12yz + 30xz and express as a perfect square.
Answer: \( (5x - 2y + 3z)^2 \)
In simple words: Identify the three base terms from the squared coefficients, verify that their cross products generate the given middle terms, and write the perfect square trinomial.

Exam Tip: Cross-check your sign choices by computing one cross product fully - if it matches, the rest will follow the same pattern.

Question 6. Evaluate (99)² and (998)² using the formula (a - b)² = a² - 2ab + b².
Answer:
(i) \( (99)^2 = (100 - 1)^2 = 10000 - 200 + 1 = 9801 \)
(ii) \( (998)^2 = (1000 - 2)^2 = 1000000 - 4000 + 4 = 996004 \)
In simple words: Rewrite each number as a difference from a nearby perfect square, apply the binomial square formula, and simplify to get the numerical answer.

Exam Tip: This algebraic method is faster and less error-prone than direct multiplication - especially useful when calculating squares of large or awkward numbers on an exam.

Question 1. Expand \( (3x + 2)^3 \)
Answer: Using the formula \( (a + b)^3 = a^3 + b^3 + 3ab(a + b) \):
\( (3x + 2)^3 = (3x)^3 + (2)^3 + 3 \times 3x \times 2 (3x + 2) \)
\( = 27x^3 + 8 + 18x(3x + 2) \)
\( = 27x^3 + 8 + 54x^2 + 36x \)

Exam Tip: Always apply the correct cube expansion formula and be careful with the coefficient multiplication at each step.

Question 2. Expand \( (3a - 2b)^3 \)
Answer: Using the formula \( (a - b)^3 = a^3 - b^3 - 3ab(a - b) \):
\( (3a - 2b)^3 = (3a)^3 - (2b)^3 - 3 \times 3a \times 2b (3a - 2b) \)
\( = 27a^3 - 8b^3 - 18ab(3a - 2b) \)
\( = 27a^3 - 8b^3 - 54a^2b + 36ab^2 \)

Exam Tip: Pay close attention to the signs - subtraction in the base leads to alternating signs in the expansion.

Question 3. Find the cube of 95.
Answer: Express 95 as a difference: \( 95 = 100 - 5 \)
\( (95)^3 = (100 - 5)^3 = (100)^3 - (5)^3 - 3 \times 100 \times 5(100 - 5) \)
\( = 1000000 - 125 - 1500(95) \)
\( = 1000000 - 125 - 142500 \)
\( = 857375 \)

Exam Tip: Breaking down a number into a difference from a perfect cube base makes calculation far simpler than direct multiplication.

Question 4. Find the cube of 999.
Answer: Express 999 as a difference: \( 999 = 1000 - 1 \)
\( (999)^3 = (1000 - 1)^3 = (1000)^3 - (1)^3 - 3 \times 1000 \times 1(1000 - 1) \)
\( = 1000000000 - 1 - 3000(999) \)
\( = 1000000000 - 1 - 2997000 \)
\( = 997002999 \)

Exam Tip: Recognising that 999 is 1 less than 1000 allows you to use the difference cube formula instead of tedious long multiplication.

Exercise 2J

Question 1. Expand \( (2x - \frac{2}{x})^3 \)
Answer: Using the formula \( (a - b)^3 = a^3 - b^3 - 3ab(a - b) \):
\( (2x - \frac{2}{x})^3 = (2x)^3 - (\frac{2}{x})^3 - 3 \times 2x \times \frac{2}{x}(2x - \frac{2}{x}) \)
\( = 8x^3 - \frac{8}{x^3} - 12(2x - \frac{2}{x}) \)
\( = 8x^3 - \frac{8}{x^3} - 24x + \frac{24}{x} \)

Exam Tip: Handle fractional terms carefully, ensuring each power of the fraction is computed accurately before simplification.

Question 2. Expand \( (3a + \frac{1}{4b})^3 \)
Answer: Using the formula \( (a + b)^3 = a^3 + b^3 + 3ab(a + b) \):
\( (3a + \frac{1}{4b})^3 = (3a)^3 + (\frac{1}{4b})^3 + 3 \times 3a \times \frac{1}{4b}(3a + \frac{1}{4b}) \)
\( = 27a^3 + \frac{1}{64b^3} + \frac{9a}{4b}(3a + \frac{1}{4b}) \)
\( = 27a^3 + \frac{1}{64b^3} + \frac{27a^2}{4b} + \frac{9a}{16b^2} \)

Exam Tip: When dealing with reciprocal terms, maintain careful control of denominators and keep intermediate results organised.

Question 3. Expand \( (\frac{4}{5}x - 2)^3 \)
Answer: Using the formula \( (a - b)^3 = a^3 - b^3 - 3ab(a - b) \):
\( (\frac{4}{5}x - 2)^3 = (\frac{4}{5}x)^3 - (2)^3 - 3 \times \frac{4}{5}x \times 2(\frac{4}{5}x - 2) \)
\( = \frac{64}{125}x^3 - 8 - \frac{24}{5}x(\frac{4}{5}x - 2) \)
\( = \frac{64}{125}x^3 - 8 - \frac{96}{25}x^2 + \frac{48}{5}x \)

Exam Tip: Simplify all fraction products before combining terms - this reduces errors and keeps your working clear.

Exercise 2K (Sum of Two Cubes)

Question 1. Factor \( x^3 + 27 \)
Answer: Rewrite as a sum of cubes: \( x^3 + 27 = x^3 + 3^3 \)
Apply the formula \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \):
\( x^3 + 27 = (x + 3)(x^2 - 3x + 9) \)

Exam Tip: Always identify perfect cubes correctly - here 27 is \( 3^3 \), which makes the factorisation straightforward.

Question 2. Factor \( 8x^3 + 27y^3 \)
Answer: Rewrite as a sum of cubes: \( 8x^3 + 27y^3 = (2x)^3 + (3y)^3 \)
Apply the formula \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \):
\( 8x^3 + 27y^3 = (2x + 3y)[(2x)^2 - (2x)(3y) + (3y)^2] \)
\( = (2x + 3y)(4x^2 - 6xy + 9y^2) \)

Exam Tip: Recognise that 8 and 27 are perfect cubes - factoring them correctly makes finding the sum of cubes pattern clear.

Question 3. Factor \( 343 + 125b^3 \)
Answer: Rewrite as a sum of cubes: \( 343 + 125b^3 = (7)^3 + (5b)^3 \)
Apply the formula \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \):
\( 343 + 125b^3 = (7 + 5b)[(7)^2 - (7)(5b) + (5b)^2] \)
\( = (7 + 5b)(49 - 35b + 25b^2) \)

Exam Tip: Check that both terms are perfect cubes before applying the formula - 343 equals \( 7^3 \) and 125 equals \( 5^3 \).

Question 4. Factor \( 1 + 64x^3 \)
Answer: Rewrite as a sum of cubes: \( 1 + 64x^3 = (1)^3 + (4x)^3 \)
Apply the formula \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \):
\( 1 + 64x^3 = (1 + 4x)[(1)^2 - 1(4x) + (4x)^2] \)
\( = (1 + 4x)(1 - 4x + 16x^2) \)

Exam Tip: Don't overlook 1 as a perfect cube - it equals \( 1^3 \) and appears in many factorisation problems.

Question 5. Factor \( 125a^3 + \frac{1}{8} \)
Answer: Rewrite as a sum of cubes: \( 125a^3 + \frac{1}{8} = (5a)^3 + (\frac{1}{2})^3 \)
Apply the formula \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \):
\( 125a^3 + \frac{1}{8} = (5a + \frac{1}{2})[(5a)^2 - 5a \times \frac{1}{2} + (\frac{1}{2})^2] \)
\( = (5a + \frac{1}{2})(25a^2 - \frac{5a}{2} + \frac{1}{4}) \)

Exam Tip: When dealing with fractions, express them as powers of rational numbers - here \( \frac{1}{8} = (\frac{1}{2})^3 \).

Question 6. Factor \( 216x^3 + \frac{1}{125} \)
Answer: Rewrite as a sum of cubes: \( 216x^3 + \frac{1}{125} = (6x)^3 + (\frac{1}{5})^3 \)
Apply the formula \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \):
\( 216x^3 + \frac{1}{125} = (6x + \frac{1}{5})[(6x)^2 - 6x \times \frac{1}{5} + (\frac{1}{5})^2] \)
\( = (6x + \frac{1}{5})(36x^2 - \frac{6x}{5} + \frac{1}{25}) \)

Exam Tip: Always verify your cube roots - 216 is \( 6^3 \) and 125 is \( 5^3 \) - this accuracy ensures correct factorisation.

Question 7. Factor \( 16x^4 + 54x \)
Answer: First, factor out the common factor: \( 16x^4 + 54x = 2x(8x^3 + 27) \)
Now recognise \( 8x^3 + 27 \) as a sum of cubes: \( 8x^3 + 27 = (2x)^3 + (3)^3 \)
Apply the formula: \( 2x[(2x)^3 + (3)^3] = 2x(2x + 3)[(2x)^2 - (2x)(3) + (3)^2] \)
\( = 2x(2x + 3)(4x^2 - 6x + 9) \)

Exam Tip: Always look for common factors first - removing them simplifies the remaining expression considerably.

Question 8. Factor \( 7a^3 + 56b^3 \)
Answer: First, factor out the common factor: \( 7a^3 + 56b^3 = 7(a^3 + 8b^3) \)
Now recognise \( a^3 + 8b^3 \) as a sum of cubes: \( a^3 + 8b^3 = (a)^3 + (2b)^3 \)
Apply the formula: \( 7[(a)^3 + (2b)^3] = 7(a + 2b)[a^2 - a(2b) + (2b)^2] \)
\( = 7(a + 2b)(a^2 - 2ab + 4b^2) \)

Exam Tip: The constant multiple in front can hide the sum of cubes pattern - always factor it out before checking for further factorisation.

Question 9. Factor \( x^5 + x^2 \)
Answer: First, factor out the common factor: \( x^5 + x^2 = x^2(x^3 + 1) \)
Now recognise \( x^3 + 1 \) as a sum of cubes: \( x^3 + 1 = (x)^3 + (1)^3 \)
Apply the formula: \( x^2[(x)^3 + (1)^3] = x^2(x + 1)[x^2 - x(1) + (1)^2] \)
\( = x^2(x + 1)(x^2 - x + 1) \)

Exam Tip: When the exponent is greater than 3, always extract the highest common factor to reveal any cubic patterns beneath.

Question 10. Factor \( a^3 + 0.008 \)
Answer: Rewrite as a sum of cubes: \( a^3 + 0.008 = (a)^3 + (0.2)^3 \)
Apply the formula \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \):
\( a^3 + 0.008 = (a + 0.2)[a^2 - a(0.2) + (0.2)^2] \)
\( = (a + 0.2)(a^2 - 0.2a + 0.04) \)

Exam Tip: Convert decimals to fractions if it helps - here 0.008 equals \( 0.2^3 \), making the sum of cubes pattern visible.

Question 11. Factor \( x^6 + y^6 \)
Answer: Rewrite as a sum of cubes: \( x^6 + y^6 = (x^2)^3 + (y^2)^3 \)
Apply the formula \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \):
\( x^6 + y^6 = (x^2 + y^2)[(x^2)^2 - x^2 y^2 + (y^2)^2] \)
\( = (x^2 + y^2)(x^4 - x^2y^2 + y^4) \)

Exam Tip: Recognise when higher powers can be expressed as cubes of lower powers - this reveals the hidden sum of cubes structure.

Question 12. Factor \( 2a^3 + 16b^3 - 5a - 10b \)
Answer: Group terms strategically: \( 2a^3 + 16b^3 - 5a - 10b = 2(a^3 + 8b^3) - 5(a + 2b) \)
Recognise the sum of cubes: \( 2[(a)^3 + (2b)^3] - 5(a + 2b) \)
\( = 2(a + 2b)[a^2 - 2ab + 4b^2] - 5(a + 2b) \)
\( = (a + 2b)[2(a^2 - 2ab + 4b^2) - 5] \)
\( = (a + 2b)(2a^2 - 4ab + 8b^2 - 5) \)

Exam Tip: When an expression has multiple terms, look for common groupings that allow factorisation of the sum of cubes alongside algebraic extraction.

Question 13. Factor \( x^3 - 512 \)
Answer: Rewrite as a difference of cubes: \( x^3 - 512 = (x)^3 - (8)^3 \)
Apply the formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \):
\( x^3 - 512 = (x - 8)[x^2 + 8x + 64] \)

Exam Tip: The difference of cubes formula uses a plus sign in the middle term of the quadratic factor - don't mix it up with the sum formula.

Question 14. Factor \( 64x^3 - 343 \)
Answer: Rewrite as a difference of cubes: \( 64x^3 - 343 = (4x)^3 - (7)^3 \)
Apply the formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \):
\( 64x^3 - 343 = (4x - 7)[(4x)^2 + (4x)(7) + (7)^2] \)
\( = (4x - 7)(16x^2 + 28x + 49) \)

Exam Tip: Both 64 and 343 are perfect cubes - recognising this immediately helps you apply the difference of cubes formula.

Question 15. Factor \( 1 - 27x^3 \)
Answer: Rewrite as a difference of cubes: \( 1 - 27x^3 = (1)^3 - (3x)^3 \)
Apply the formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \):
\( 1 - 27x^3 = (1 - 3x)[1 + 3x + 9x^2] \)

Exam Tip: Even though 1 seems simple, always treat it as \( 1^3 \) when applying cube factorisation formulas.

Question 16. Factor \( 1 - 27x^3 \)
Answer: Rewrite as a difference of cubes: \( 1 - 27x^3 = (1)^3 - (3x)^3 \)
Apply the formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \):
\( 1 - 27x^3 = (1 - 3x)[1 + 3x + 9x^2] \)

Exam Tip: This is the same as Question 15 - always verify that your final answer is correct by expanding to check.

Question 17. Factor \( a^3 - 0.064 \)
Answer: Rewrite as a difference of cubes: \( a^3 - 0.064 = (a)^3 - (0.4)^3 \)
Apply the formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \):
\( a^3 - 0.064 = (a - 0.4)[a^2 + 0.4a + 0.16] \)

Exam Tip: Convert decimal values carefully - here 0.064 equals \( (0.4)^3 \) and 0.4 equals \( \frac{2}{5} \).

Question 18. Factor \( (a + b)^3 - 8 \)
Answer: Rewrite as a difference of cubes: \( (a + b)^3 - 8 = (a + b)^3 - (2)^3 \)
Apply the formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \):
\( (a + b)^3 - 8 = [(a + b) - 2][(a + b)^2 + (a + b)(2) + (2)^2] \)
\( = (a + b - 2)[a^2 + b^2 + 2ab + 2(a + b) + 4] \)

Exam Tip: When the first term is itself a power expression, treat the entire expression as a single unit for the difference of cubes formula.

Question 19. Factor \( x^6 - 729 \)
Answer: Rewrite as a difference of cubes: \( x^6 - 729 = (x^2)^3 - (9)^3 \)
Apply the formula: \( (x^2 - 9)[(x^2)^2 + x^2 \times 9 + (9)^2] \)
\( = (x^2 - 9)(x^4 + 9x^2 + 81) \)
Further factor \( x^2 - 9 \) as a difference of squares: \( (x + 3)(x - 3)(x^4 + 9x^2 + 81) \)
Check if \( x^4 + 9x^2 + 81 \) factors further using the pattern for \( (x^2)^2 + 3(x^2) + 9 - (3x)^2 \):
\( = (x + 3)(x - 3)(x^2 + 3x + 9)(x^2 - 3x + 9) \)

Exam Tip: After the first factorisation step, check whether the resulting quadratic or quartic can be factored further - this often reveals hidden patterns.

Question 20. Factor \( (a + b)^3 - (a - b)^3 \)
Answer: Using the difference of cubes formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \), set \( a = (a+b) \) and \( b = (a-b) \):
\( [(a + b) - (a - b)][(a + b)^2 + (a + b)(a - b) + (a - b)^2] \)
\( = [2b][a^2 + 2ab + b^2 + a^2 - b^2 + a^2 - 2ab + b^2] \)
\( = 2b[3a^2 + b^2] \)

Exam Tip: When applying difference of cubes to grouped expressions, expand all squares carefully to avoid sign errors.

Question 21. Factor \( x^3 - 8xy^3 \)
Answer: First extract the common factor: \( x^3 - 8xy^3 = x(1 - 8y^3) \)
Now recognise \( 1 - 8y^3 \) as a difference of cubes: \( 1 - 8y^3 = (1)^3 - (2y)^3 \)
Apply the formula: \( x[(1 - 2y)(1 + 2y + 4y^2)] \)
\( = x(1 - 2y)(1 + 2y + 4y^2) \)

Exam Tip: Always extract common factors before attempting any other factorisation method - it simplifies the remaining expression.

Question 22. Factor \( 32x^4 - 500x \)
Answer: First extract the common factor: \( 32x^4 - 500x = 4x(8x^3 - 125) \)
Now recognise \( 8x^3 - 125 \) as a difference of cubes: \( 8x^3 - 125 = (2x)^3 - (5)^3 \)
Apply the formula: \( 4x[(2x - 5)(4x^2 + 10x + 25)] \)
\( = 4x(2x - 5)(4x^2 + 10x + 25) \)

Exam Tip: Factor out the greatest common factor systematically - both the coefficient and all variable terms should be considered.

Question 23. Factor \( 8x^3 - \frac{1}{27y^3} \)
Answer: Rewrite as a difference of cubes: \( 8x^3 - \frac{1}{27y^3} = (2x)^3 - (\frac{1}{3y})^3 \)
Apply the formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \):
\( (2x - \frac{1}{3y})[(2x)^2 + 2x \times \frac{1}{3y} + (\frac{1}{3y})^2] \)
\( = (2x - \frac{1}{3y})(4x^2 + \frac{2x}{3y} + \frac{1}{9y^2}) \)

Exam Tip: When denominators are present, convert them to fractional exponents to identify the cube root structure clearly.

Exercise 2K (continued)

Question 24. Factor \( 3a^7b - 81a^4b^4 \)
Answer: First extract the common factor: \( 3a^7b - 81a^4b^4 = 3a^4b(a^3 - 27b^3) \)
Now recognise \( a^3 - 27b^3 \) as a difference of cubes: \( a^3 - 27b^3 = (a)^3 - (3b)^3 \)
Apply the formula: \( 3a^4b(a - 3b)[a^2 + a(3b) + (3b)^2] \)
\( = 3a^4b(a - 3b)(a^2 + 3ab + 9b^2) \)

Exam Tip: When extracting common factors from terms with higher powers, write out the lowest power for each variable to avoid errors.

Question 25. Prove that \( a^3 - \frac{1}{a^3} = 2a - \frac{2}{a} \) when \( a - \frac{1}{a} = 2 \)
Answer: We know that \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)
\( a^3 - \frac{1}{a^3} = (a - \frac{1}{a})(a^2 + a \times \frac{1}{a} + \frac{1}{a^2}) \)
\( = (a - \frac{1}{a})(a^2 + 1 + \frac{1}{a^2}) \)
From \( a - \frac{1}{a} = 2 \), we get \( (a - \frac{1}{a})^2 = 4 \)
\( a^2 - 2 + \frac{1}{a^2} = 4 \)
\( a^2 + \frac{1}{a^2} = 6 \)
Therefore: \( (a - \frac{1}{a})(a^2 + 1 + \frac{1}{a^2}) = (a - \frac{1}{a})(6 + 1) = 7(a - \frac{1}{a}) = 7 \times 2 = 14 \)
\( a^3 - \frac{1}{a^3} = (a - \frac{1}{a})(a^2 + 1 + \frac{1}{a^2}) = 2(a^2 + 1 + \frac{1}{a^2}) \)
\( = (a - \frac{1}{a})^2 + 2 + 2 = 4 + 4 = 2(a - \frac{1}{a}) + 2 = 2a - \frac{2}{a} + 2 \)

Exam Tip: When given a constraint equation, manipulate it algebraically to find useful relationships like squares or sums of powers before substituting into the main expression.

Question 26. Factor \( 8a^3 - b^3 - 4ax + 2bx \)
Answer: Group the first two terms and the last two terms: \( (8a^3 - b^3) - (4ax - 2bx) \)
Factor the difference of cubes and extract the common factor: \( (2a)^3 - (b)^3 - 2x(2a - b) \)
\( = (2a - b)[(2a)^2 + 2a(b) + (b)^2] - 2x(2a - b) \)
\( = (2a - b)[4a^2 + 2ab + b^2] - 2x(2a - b) \)
\( = (2a - b)[4a^2 + 2ab + b^2 - 2x] \)

Exam Tip: When an expression mixes cubes and linear terms, group strategically so that a common binomial emerges for final factorisation.

Question 27. Factor \( 8a^3 - b^3 - 4ax + 2bx \)
Answer: Group strategically: \( 8a^3 - b^3 - 2x(2a - b) \)
Factor the difference of cubes: \( (2a - b)[(2a)^2 + 2a(b) + b^2] - 2x(2a - b) \)
\( = (2a - b)[4a^2 + 2ab + b^2] - 2x(2a - b) \)
Extract the common binomial: \( (2a - b)(4a^2 + 2ab + b^2 - 2x) \)

Exam Tip: This repeats Question 26 - double-check by expanding your factored form to verify correctness.