ML Aggarwal Class 10 Maths Solutions Chapter 06 Factorization

Question 1. Find the remainder (without division) on dividing f(x) by (x - 2) where
(i) f(x) = 5x² - 7x + 4
(ii) f(x) = 2x³ - 7x² + 3
Answer: Using the remainder theorem, when f(x) is divided by (x - a), the remainder equals f(a).

(i) When dividing f(x) = 5x² - 7x + 4 by (x - 2):

Remainder = f(2) = 5(2)² - 7(2) + 4 = 5(4) - 14 + 4 = 20 - 14 + 4 = 10

(ii) When dividing f(x) = 2x³ - 7x² + 3 by (x - 2):

Remainder = f(2) = 2(2)³ - 7(2)² + 3 = 2(8) - 7(4) + 3 = 16 - 28 + 3 = -9
In simple words: Substitute x = 2 (the value that makes x - 2 equal to 0) into the polynomial. The result you get is the remainder.

Exam Tip: Always identify the value of x that makes the divisor zero - that's what you substitute into f(x) to find the remainder quickly without doing long division.

Question 2. Using remainder theorem, find the remainder on dividing f(x) by (x + 3) where
(i) f(x) = 2x² - 5x + 1
(ii) f(x) = 3x³ + 7x² - 5x + 1
Answer: By the remainder theorem, when f(x) is divided by (x - a), the remainder is f(a).

(i) When dividing f(x) = 2x² - 5x + 1 by (x + 3) or (x - (-3)):

Remainder = f(-3) = 2(-3)² - 5(-3) + 1 = 2(9) + 15 + 1 = 18 + 15 + 1 = 34

(ii) When dividing f(x) = 3x³ + 7x² - 5x + 1 by (x + 3) or (x - (-3)):

Remainder = f(-3) = 3(-3)³ + 7(-3)² - 5(-3) + 1 = 3(-27) + 7(9) + 15 + 1 = -81 + 63 + 15 + 1 = -2
In simple words: Since the divisor is (x + 3), put x = -3 into the polynomial and calculate to get the remainder.

Exam Tip: Watch the sign carefully - when the divisor is (x + 3), you substitute x = -3, not x = 3. This is a common source of errors.

Question 3. Find the remainder (without division) on dividing f(x) by (2x + 1) where
(i) f(x) = 4x² + 5x + 3
(ii) f(x) = 3x³ - 7x² + 4x + 11
Answer: Using the remainder theorem, when f(x) is divided by (x - a), the remainder is f(a).

(i) When dividing f(x) = 4x² + 5x + 3 by (2x + 1) or \( 2\left(x - \left(-\frac{1}{2}\right)\right) \):

Remainder = \( f\left(-\frac{1}{2}\right) = 4\left(-\frac{1}{2}\right)^2 + 5\left(-\frac{1}{2}\right) + 3 = 4\left(\frac{1}{4}\right) - \frac{5}{2} + 3 = 1 - \frac{5}{2} + 3 \)

Taking LCM: \( = \frac{2 - 5 + 6}{2} = \frac{3}{2} = 1\frac{1}{2} \)

(ii) When dividing f(x) = 3x³ - 7x² + 4x + 11 by (2x + 1) or \( 2\left(x - \left(-\frac{1}{2}\right)\right) \):

Remainder = \( f\left(-\frac{1}{2}\right) = 3\left(-\frac{1}{2}\right)^3 - 7\left(-\frac{1}{2}\right)^2 + 4\left(-\frac{1}{2}\right) + 11 = 3\left(-\frac{1}{8}\right) - 7\left(\frac{1}{4}\right) - 2 + 11 \)

\( = -\frac{3}{8} - \frac{7}{4} + 9 = \frac{-3 - 14 + 72}{8} = \frac{55}{8} = 6\frac{7}{8} \)
In simple words: When the divisor has a coefficient other than 1 (like 2x + 1), first rewrite it as a multiple, then set it equal to zero to find the value to substitute.

Exam Tip: For divisors like (2x + 1) or (3x - 2), remember to find the exact value by solving 2x + 1 = 0, which gives fractional values. Use these fractions when substituting.

Question 4. Using remainder theorem, find the value of k if on dividing 2x³ + 3x² - kx + 5 by (x - 2) leaves a remainder 7.
Answer: By the remainder theorem, when f(x) is divided by (x - a), the remainder equals f(a).

When dividing f(x) = 2x³ + 3x² - kx + 5 by (x - 2):

Remainder = f(2) = 2(2)³ + 3(2)² - k(2) + 5

Given that remainder = 7, we have:

\( 2(8) + 3(4) - 2k + 5 = 7 \)

\( 16 + 12 - 2k + 5 = 7 \)

\( 33 - 2k = 7 \)

\( 2k = 33 - 7 = 26 \)

\( k = 13 \)
In simple words: Calculate f(2), set it equal to 7, and solve for k by rearranging the equation.

Exam Tip: Always substitute the value first to get an expression with the unknown variable, then treat it as a simple equation to solve.

Question 5. Using remainder theorem, find the value of a if the division of x³ + 5x² - ax + 6 by (x - 1) leaves the remainder 2a.
Answer: By the remainder theorem, when f(x) is divided by (x - a), the remainder is f(a).

When dividing f(x) = x³ + 5x² - ax + 6 by (x - 1):

Remainder = f(1) = (1)³ + 5(1)² - a(1) + 6 = 1 + 5 - a + 6 = 12 - a

Given that remainder = 2a:

\( 12 - a = 2a \)

\( 12 = 3a \)

\( a = 4 \)
In simple words: Substitute x = 1, get an expression in terms of a, then set it equal to 2a and solve for a.

Exam Tip: When the remainder is given in terms of an unknown (like 2a), set your calculated remainder equal to that expression to form an equation.

Question 6(i). What number must be subtracted from 2x² - 5x so that resulting polynomial leaves remainder 2 when divided by 2x + 1?
Answer: Let the number to be subtracted be a.

The new polynomial becomes: 2x² - 5x - a

By the remainder theorem, when this polynomial is divided by (2x + 1) or \( 2\left(x - \left(-\frac{1}{2}\right)\right) \):

Remainder = \( f\left(-\frac{1}{2}\right) = 2\left(-\frac{1}{2}\right)^2 - 5\left(-\frac{1}{2}\right) - a = 2\left(\frac{1}{4}\right) + \frac{5}{2} - a = \frac{1}{2} + \frac{5}{2} - a = 3 - a \)

Given that remainder = 2:

\( 3 - a = 2 \)

\( a = 1 \)
In simple words: Subtract an unknown number from the polynomial, then use the remainder theorem to find what that number must be.

Exam Tip: Set up the polynomial with the unknown subtracted, apply the remainder theorem, then solve for the unknown by equating it to the given remainder.

Question 6(ii). What number must be added to 2x³ - 3x² - 8x so that resulting polynomial leaves the remainder 10 when divided by 2x + 1?
Answer: From the divisor 2x + 1 = 0, we get \( x = -\frac{1}{2} \)

Let the number added be a.

The new polynomial is: 2x³ - 3x² - 8x + a

By the remainder theorem:

\( f\left(-\frac{1}{2}\right) = 2\left(-\frac{1}{2}\right)^3 - 3\left(-\frac{1}{2}\right)^2 - 8\left(-\frac{1}{2}\right) + a \)

\( = 2\left(-\frac{1}{8}\right) - 3\left(\frac{1}{4}\right) + 4 + a = -\frac{1}{4} - \frac{3}{4} + 4 + a = -1 + 4 + a = 3 + a \)

Given that remainder = 10:

\( 3 + a = 10 \)

\( a = 7 \)
In simple words: Add an unknown to the polynomial, apply the remainder theorem at \( x = -\frac{1}{2} \), then solve for the unknown.

Exam Tip: When adding or subtracting from a polynomial, always apply the remainder theorem to find the unknown, not long division.

Question 7(i). When divided by x - 3 the polynomials x³ - px² + x + 6 and 2x³ - x² - (p + 3)x - 6 leave the same remainder. Find the value of 'p'.
Answer: By the remainder theorem, when f(x) is divided by (x - b), the remainder is f(b).

For the first polynomial f(x) = x³ - px² + x + 6 divided by (x - 3):

Remainder = f(3) = (3)³ - p(3)² + 3 + 6 = 27 - 9p + 9 = 36 - 9p

For the second polynomial g(x) = 2x³ - x² - (p + 3)x - 6 divided by (x - 3):

Remainder = g(3) = 2(3)³ - (3)² - (p + 3)(3) - 6 = 2(27) - 9 - 3p - 9 - 6 = 54 - 9 - 3p - 9 - 6 = 30 - 3p

Since both remainders are equal:

\( 36 - 9p = 30 - 3p \)

\( 36 - 30 = 9p - 3p \)

\( 6 = 6p \)

\( p = 1 \)
In simple words: Find the remainder for each polynomial at x = 3, then set them equal to solve for p.

Exam Tip: When two polynomials leave the same remainder, calculate both remainders separately and equate them to find the unknown parameter.

Question 7(ii). Find 'a' if the two polynomials ax³ + 3x² - 9 and 2x³ + 4x + a, leaves the same remainder when divided by x + 3.
Answer: By the remainder theorem, when f(x) is divided by (x - b), the remainder is f(b).

For the first polynomial f(x) = ax³ + 3x² - 9 divided by (x + 3) or (x - (-3)):

Remainder = f(-3) = a(-3)³ + 3(-3)² - 9 = -27a + 27 - 9 = 18 - 27a

For the second polynomial g(x) = 2x³ + 4x + a divided by (x + 3) or (x - (-3)):

Remainder = g(-3) = 2(-3)³ + 4(-3) + a = 2(-27) - 12 + a = -54 - 12 + a = a - 66

Since both remainders are equal:

\( 18 - 27a = a - 66 \)

\( 18 + 66 = a + 27a \)

\( 84 = 28a \)

\( a = 3 \)
In simple words: Calculate each remainder at x = -3, set them equal, and solve for a.

Exam Tip: Always substitute the correct value (from x + 3 = 0, so x = -3) and be careful with negative numbers when cubing.

Question 7(iii). The polynomials ax³ + 3x² - 3 and 2x³ - 5x + a when divided by x - 4 leave the remainder r₁ and r₂ respectively. If 2r₁ = r₂, then find the value of a.
Answer: By the remainder theorem, when f(x) is divided by (x - b), the remainder is f(b).

For the first polynomial f(x) = ax³ + 3x² - 3 divided by (x - 4):

r₁ = f(4) = a(4)³ + 3(4)² - 3 = 64a + 48 - 3 = 64a + 45

For the second polynomial g(x) = 2x³ - 5x + a divided by (x - 4):

r₂ = g(4) = 2(4)³ - 5(4) + a = 2(64) - 20 + a = 128 - 20 + a = 108 + a

Given that 2r₁ = r₂:

\( 2(64a + 45) = 108 + a \)

\( 128a + 90 = 108 + a \)

\( 128a - a = 108 - 90 \)

\( 127a = 18 \)

\( a = \frac{18}{127} \)
In simple words: Find r₁ and r₂ at x = 4, then use the condition 2r₁ = r₂ to set up an equation for a.

Exam Tip: When a relationship is given between two remainders (like 2r₁ = r₂ or r₁ + r₂ = constant), substitute the expressions and solve carefully.

Question 8. Using the remainder theorem, find the remainders obtained when x³ + (kx + 8)x + k is divided by x + 1 and x - 2. Hence, find k if the sum of two remainders is 1.
Answer: By the remainder theorem, when f(x) is divided by (x - b), the remainder is f(b).

Let f(x) = x³ + (kx + 8)x + k = x³ + kx² + 8x + k

When dividing by (x + 1) or (x - (-1)):

r₁ = f(-1) = (-1)³ + k(-1)² + 8(-1) + k = -1 + k - 8 + k = 2k - 9

When dividing by (x - 2):

r₂ = f(2) = (2)³ + k(2)² + 8(2) + k = 8 + 4k + 16 + k = 5k + 24

Given that the sum of the two remainders = 1:

\( r₁ + r₂ = 1 \)

\( (2k - 9) + (5k + 24) = 1 \)

\( 7k + 15 = 1 \)

\( 7k = -14 \)

\( k = -2 \)
In simple words: Find the remainder at x = -1 and x = 2, add them together, and set equal to 1 to solve for k.

Exam Tip: Always expand the polynomial fully first, then substitute the values to find remainders. This prevents calculation errors.

Question 9. By factor theorem, show that (x + 3) and (2x - 1) are the factors of 2x² + 5x - 3.
Answer: By the factor theorem, (x - a) is a factor of f(x) if f(a) = 0.

Let f(x) = 2x² + 5x - 3

For (x + 3) = (x - (-3)) to be a factor, we check if f(-3) = 0:

f(-3) = 2(-3)² + 5(-3) - 3 = 2(9) - 15 - 3 = 18 - 15 - 3 = 0 ✓

For (2x - 1) = 2(x - 1/2) to be a factor, we check if f(1/2) = 0:

\( f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^2 + 5\left(\frac{1}{2}\right) - 3 = 2\left(\frac{1}{4}\right) + \frac{5}{2} - 3 = \frac{1}{2} + \frac{5}{2} - 3 = 3 - 3 = 0 \) ✓

Since f(-3) = 0 and f(1/2) = 0, both (x + 3) and (2x - 1) are factors of 2x² + 5x - 3.
In simple words: To prove something is a factor, substitute the value that makes it zero into the polynomial and show the result is zero.

Exam Tip: For a factor like (2x - 1), solve 2x - 1 = 0 to get x = 1/2, then use this exact value when substituting.

Question 10. Without actual division, prove that x⁴ + 2x³ - 2x² + 2x - 3 is exactly divisible by x² + 2x - 3.
Answer: Let f(x) = x⁴ + 2x³ - 2x² + 2x - 3 and g(x) = x² + 2x - 3

First, factorise g(x):

g(x) = x² + 2x - 3 = x² + 3x - x - 3 = x(x + 3) - 1(x + 3) = (x - 1)(x + 3)

For f(x) to be exactly divisible by g(x), both (x - 1) and (x + 3) must be factors of f(x), meaning f(1) = 0 and f(-3) = 0.

Check f(1):

f(1) = (1)⁴ + 2(1)³ - 2(1)² + 2(1) - 3 = 1 + 2 - 2 + 2 - 3 = 0 ✓

Check f(-3):

f(-3) = (-3)⁴ + 2(-3)³ - 2(-3)² + 2(-3) - 3 = 81 + 2(-27) - 2(9) - 6 - 3 = 81 - 54 - 18 - 6 - 3 = 0 ✓

Since f(1) = 0 and f(-3) = 0, both (x - 1) and (x + 3) are factors of f(x). Therefore, f(x) is exactly divisible by g(x) = (x - 1)(x + 3) = x² + 2x - 3.
In simple words: Factorise the divisor, then show that both its factors are also factors of the dividend by substituting their zeros.

Exam Tip: To prove divisibility without division, factor the divisor and verify that both factors of the divisor are also factors of the dividend.

Question 11. Show that (x - 2) is a factor 3x² - x - 10. Hence, factorize 3x² - x - 10.
Answer: By the factor theorem, (x - a) is a factor of f(x) if f(a) = 0.

Let f(x) = 3x² - x - 10

Check if (x - 2) is a factor by verifying f(2) = 0:

f(2) = 3(2)² - 2 - 10 = 3(4) - 2 - 10 = 12 - 2 - 10 = 0 ✓

Hence, (x - 2) is a factor of 3x² - x - 10.

Now, to factorise 3x² - x - 10:

\( 3x^2 - x - 10 = 3x^2 - 6x + 5x - 10 = 3x(x - 2) + 5(x - 2) = (3x + 5)(x - 2) \)

Therefore, 3x² - x - 10 = (x - 2)(3x + 5)
In simple words: Verify (x - 2) is a factor, then split the middle term to complete the factorisation.

Exam Tip: After proving a factor, use the factorisation method by splitting the middle term in a way that gives you the known factor.

Question 12. Using factor theorem, show that (x - 2) is a factor of x³ + x² - 4x - 4. Hence, factorise the polynomial completely.
Answer: By the factor theorem, (x - a) is a factor of f(x) if f(a) = 0.

Let f(x) = x³ + x² - 4x - 4

Check if (x - 2) is a factor by verifying f(2) = 0:

f(2) = (2)³ + (2)² - 4(2) - 4 = 8 + 4 - 8 - 4 = 0 ✓

Hence, (x - 2) is a factor of x³ + x² - 4x - 4.

Now, to factorise x³ + x² - 4x - 4 completely:

\( x^3 + x^2 - 4x - 4 = x^2(x + 1) - 4(x + 1) = (x^2 - 4)(x + 1) = (x - 2)(x + 2)(x + 1) \)

Therefore, x³ + x² - 4x - 4 = (x - 2)(x + 1)(x + 2)
In simple words: Verify the given factor, then group terms in the polynomial to find the other factors through factorisation.

Exam Tip: After confirming one factor, use grouping method to find the remaining factors rather than doing polynomial division.

Question 13. Show that 2x + 7 is a factor of 2x³ + 5x² - 11x - 14. Hence, factorise the given expression completely, using the factor theorem.
Answer: By the factor theorem, (x - a) is a factor of f(x) if f(a) = 0.

Let f(x) = 2x³ + 5x² - 11x - 14

(2x + 7) = 2(x - (-7/2)) is a factor if f(-7/2) = 0:

\( f\left(-\frac{7}{2}\right) = 2\left(-\frac{7}{2}\right)^3 + 5\left(-\frac{7}{2}\right)^2 - 11\left(-\frac{7}{2}\right) - 14 \)

\( = 2\left(-\frac{343}{8}\right) + 5\left(\frac{49}{4}\right) + \frac{77}{2} - 14 = -\frac{343}{4} + \frac{245}{4} + \frac{154}{4} - \frac{56}{4} = \frac{-343 + 245 + 154 - 56}{4} = \frac{0}{4} = 0 \) ✓

Hence, (2x + 7) is a factor of 2x³ + 5x² - 11x - 14.

Dividing f(x) by (2x + 7) gives the quotient x² - x - 2, which factors as:

\( x^2 - x - 2 = (x - 2)(x + 1) \)

Therefore, 2x³ + 5x² - 11x - 14 = (2x + 7)(x + 1)(x - 2)
In simple words: Verify (2x + 7) is a factor by substituting x = -7/2, then divide to find the quotient and factorise it further.

Exam Tip: For divisors with coefficients like 2x + 7, solve 2x + 7 = 0 carefully to get the exact value x = -7/2 to substitute.

Question 14. Use factor theorem to factorise the following polynomials completely:
(i) x³ + 2x² - 5x - 6
(ii) x³ - 13x - 12
(iii) 6x³ + 17x² + 4x - 12
Answer:

(i) Let f(x) = x³ + 2x² - 5x - 6

Testing small values: f(1) = 1 + 2 - 5 - 6 = -8, f(-1) = -1 + 2 + 5 - 6 = 0

So (x + 1) is a factor. Dividing f(x) by (x + 1) gives x² + x - 6, which factors as (x + 3)(x - 2).

Therefore, x³ + 2x² - 5x - 6 = (x + 1)(x + 3)(x - 2)

(ii) Let f(x) = x³ - 13x - 12

Testing small values: f(-1) = -1 + 13 - 12 = 0

So (x + 1) is a factor. Dividing f(x) by (x + 1) gives x² - x - 12, which factors as (x - 4)(x + 3).

Therefore, x³ - 13x - 12 = (x + 1)(x - 4)(x + 3)

(iii) Let f(x) = 6x³ + 17x² + 4x - 12

Testing fractional values: f(2/3) = 6(8/27) + 17(4/9) + 4(2/3) - 12 = 16/9 + 68/9 + 8/3 - 12 = 0

So (3x - 2) is a factor. Dividing f(x) by (3x - 2) gives 2x² + 7x + 6, which factors as (2x + 3)(x + 2).

Therefore, 6x³ + 17x² + 4x - 12 = (3x - 2)(2x + 3)(x + 2)
In simple words: Test small integer and simple fractional values to find one factor, then divide to get a quadratic, which you can factorise further.

Exam Tip: Try \( \pm1, \pm2, \pm3 \) first, then if those don't work, try simple fractions like \( \pm\frac{1}{2}, \pm\frac{1}{3}, \pm\frac{2}{3} \). For divisors like (3x - 2), solve 3x - 2 = 0 to get x = 2/3.

Question 15. Use Remainder Theorem to factorise the following polynomials completely:
(i) \( 2x^3 + x^2 - 13x + 6 \)
(ii) \( 3x^3 + 2x^2 - 19x + 6 \)
(iii) \( 2x^3 + 3x^2 - 9x - 10 \)
(iv) \( x^3 + 10x^2 - 37x + 26 \)
Answer:
(i) Let \( f(x) = 2x^3 + x^2 - 13x + 6 \). By substituting \( x = 2 \), we find \( f(2) = 0 \). Using the factor theorem, \( (x - 2) \) is a factor. Dividing \( f(x) \) by \( (x - 2) \) yields the quotient \( 2x^2 + 5x - 3 \) with remainder zero. Factoring the quadratic: \( 2x^2 + 5x - 3 = (2x - 1)(x + 3) \). Therefore, \( 2x^3 + x^2 - 13x + 6 = (x - 2)(2x - 1)(x + 3) \).
(ii) Let \( f(x) = 3x^3 + 2x^2 - 19x + 6 \). Testing \( x = 2 \) gives \( f(2) = 0 \), so \( (x - 2) \) is a factor. Dividing yields \( 3x^2 + 8x - 3 \). Factoring: \( 3x^2 + 8x - 3 = (3x - 1)(x + 3) \). Thus, \( 3x^3 + 2x^2 - 19x + 6 = (x - 2)(3x - 1)(x + 3) \).
(iii) Let \( f(x) = 2x^3 + 3x^2 - 9x - 10 \). Since \( f(2) = 0 \), \( (x - 2) \) is a factor. Division gives quotient \( 2x^2 + 7x + 5 \). Factoring: \( 2x^2 + 7x + 5 = (2x + 5)(x + 1) \). Therefore, \( 2x^3 + 3x^2 - 9x - 10 = (x - 2)(x + 1)(2x + 5) \).
(iv) Let \( f(x) = x^3 + 10x^2 - 37x + 26 \). Testing \( x = 1 \) gives \( f(1) = 0 \), making \( (x - 1) \) a factor. Division produces \( x^2 + 11x - 26 \). Factoring: \( x^2 + 11x - 26 = (x - 2)(x + 13) \). Hence, \( x^3 + 10x^2 - 37x + 26 = (x - 1)(x - 2)(x + 13) \).
In simple words: To factor a polynomial, test values until you find one that makes \( f(x) = 0 \). That tells you one factor. Then divide the polynomial by that factor and factor the result.

Exam Tip: Always use the factor theorem to find the first factor by testing small integer values, then perform polynomial division carefully to avoid arithmetic errors.

Question 16. If (2x + 1) is a factor of \( 6x^3 + 5x^2 + ax - 2 \), find the value of a.
Answer: Let \( f(x) = 6x^3 + 5x^2 + ax - 2 \). If \( (2x + 1) \) is a factor, then \( f\left(-\frac{1}{2}\right) = 0 \) by the factor theorem. Substituting: \( 6\left(-\frac{1}{2}\right)^3 + 5\left(-\frac{1}{2}\right)^2 + a\left(-\frac{1}{2}\right) - 2 = 0 \). This simplifies to \( -\frac{3}{4} + \frac{5}{4} - \frac{a}{2} - 2 = 0 \). Combining fractions: \( \frac{-3 + 5 - 2a - 8}{4} = 0 \), which gives \( -6 - 2a = 0 \). Solving: \( a = -3 \).
In simple words: If a factor is \( (2x + 1) \), then \( x = -\frac{1}{2} \) makes the polynomial equal zero. Plug in this value and solve for \( a \).

Exam Tip: When given a linear factor like \( (2x + 1) \), set it to zero to find the zero of the polynomial, then substitute into the original expression to find unknown coefficients.

Question 17. If (3x - 2) is a factor of \( 3x^3 - kx^2 + 21x - 10 \), find the value of k.
Answer: Let \( f(x) = 3x^3 - kx^2 + 21x - 10 \). Since \( (3x - 2) \) is a factor, \( f\left(\frac{2}{3}\right) = 0 \). Evaluating: \( 3\left(\frac{2}{3}\right)^3 - k\left(\frac{2}{3}\right)^2 + 21\left(\frac{2}{3}\right) - 10 = 0 \). This becomes \( \frac{8}{9} - \frac{4k}{9} + 14 - 10 = 0 \). Simplifying: \( \frac{8 - 4k + 36}{9} = 0 \), which gives \( 44 - 4k = 0 \). Therefore, \( k = 11 \).
In simple words: Set the factor \( 3x - 2 \) equal to zero to get \( x = \frac{2}{3} \). Substitute this value into the polynomial and solve for the unknown constant.

Exam Tip: Convert the factor to its zero form carefully when the coefficient of \( x \) is not 1, then substitute and solve systematically.

Question 18. If (x - 2) is a factor of \( 2x^3 - x^2 - px - 2 \), then
(i) Find the value of p.
(ii) With this value of p, factorise the above expression completely.
Answer:
(i) Let \( f(x) = 2x^3 - x^2 - px - 2 \). Since \( (x - 2) \) is a factor, \( f(2) = 0 \). Substituting: \( 2(8) - 4 - 2p - 2 = 0 \), which simplifies to \( 10 - 2p = 0 \). Therefore, \( p = 5 \).
(ii) With \( p = 5 \), we have \( f(x) = 2x^3 - x^2 - 5x - 2 \). Dividing by \( (x - 2) \) gives quotient \( 2x^2 + 3x + 1 \). Factoring the quadratic: \( 2x^2 + 3x + 1 = (2x + 1)(x + 1) \). Complete factorisation: \( 2x^3 - x^2 - 5x - 2 = (x - 2)(2x + 1)(x + 1) \).
In simple words: First, use the factor condition to find the missing constant. Then divide by the known factor and factor the remaining quadratic expression.

Exam Tip: After finding the unknown constant, verify your quadratic factorisation by expanding it back to confirm the original polynomial.

Question 19. What number should be subtracted from \( 2x^3 - 5x^2 + 5x \) so that the resulting polynomial has \( 2x - 3 \) as a factor?
Answer: Let the number to be subtracted be \( a \). The new polynomial is \( f(x) = 2x^3 - 5x^2 + 5x - a \). For \( (2x - 3) \) to be a factor, \( f\left(\frac{3}{2}\right) = 0 \). Evaluating: \( 2\left(\frac{3}{2}\right)^3 - 5\left(\frac{3}{2}\right)^2 + 5\left(\frac{3}{2}\right) - a = 0 \). This gives \( \frac{27}{4} - \frac{45}{4} + \frac{15}{2} - a = 0 \). Simplifying: \( \frac{27 - 45 + 30 - 4a}{4} = 0 \), which yields \( 12 - 4a = 0 \). Therefore, \( a = 3 \).
In simple words: If a factor is \( 2x - 3 \), then \( x = \frac{3}{2} \) makes the new polynomial zero. Substitute this into the expression with \( a \) and solve for \( a \).

Exam Tip: Always substitute the zero of the factor into the modified polynomial expression to find the unknown value you need to subtract.

Question 20(i). Find the value of the constants a and b, if (x - 2) and (x + 3) are both factors of the expression \( x^3 + ax^2 + bx - 12 \).
Answer: Let \( f(x) = x^3 + ax^2 + bx - 12 \). Since both \( (x - 2) \) and \( (x + 3) \) are factors, \( f(2) = 0 \) and \( f(-3) = 0 \). From \( f(2) = 0 \): \( 8 + 4a + 2b - 12 = 0 \), giving \( 4a + 2b = 4 \), or \( 2a + b = 2 \), so \( b = 2 - 2a \) (Equation 1). From \( f(-3) = 0 \): \( -27 + 9a - 3b - 12 = 0 \), giving \( 9a - 3b = 39 \). Substituting Equation 1: \( 9a - 3(2 - 2a) = 39 \), which simplifies to \( 15a = 45 \), so \( a = 3 \). Then \( b = 2 - 6 = -4 \).
In simple words: When you have two factors, each one gives you a separate zero. Plug both zeros into the polynomial to get two equations with two unknowns, then solve the system.

Exam Tip: Set up two separate equations from the two factor conditions, solve them as a system of equations, and always verify by substituting back into both conditions.

Question 20(ii). If (x + 2) and (x + 3) are factors of \( x^3 + ax + b \), find the values of a and b.
Answer: Let \( f(x) = x^3 + ax + b \). Since both \( (x + 2) \) and \( (x + 3) \) are factors, \( f(-2) = 0 \) and \( f(-3) = 0 \). From \( f(-2) = 0 \): \( -8 - 2a + b = 0 \), giving \( b = 2a + 8 \) (Equation 1). From \( f(-3) = 0 \): \( -27 - 3a + b = 0 \). Substituting Equation 1: \( -27 - 3a + 2a + 8 = 0 \), which simplifies to \( -a = 19 \), so \( a = -19 \). Then \( b = 2(-19) + 8 = -30 \).
In simple words: With two factors, you get two zeros. Set up two equations using these zeros, substitute one into the other, and solve for both unknowns.

Exam Tip: When a polynomial is missing a term (here, the \( x^2 \) term), be careful with the substitution - note the missing term is treated as having a coefficient of zero.

Question 21. If (x + 2) and (x - 3) are the factors of \( x^3 + ax + b \), find the values of a and b. With these values of a and b, factorise the given expression.
Answer: Let \( f(x) = x^3 + ax + b \). From \( f(-2) = 0 \): \( -8 - 2a + b = 0 \), so \( b = 2a + 8 \) (Equation 1). From \( f(3) = 0 \): \( 27 + 3a + b = 0 \). Substituting Equation 1: \( 27 + 3a + 2a + 8 = 0 \), giving \( 5a = -35 \), so \( a = -7 \). Then \( b = 2(-7) + 8 = -6 \). Thus, \( f(x) = x^3 - 7x - 6 \). Since \( (x + 2) \) and \( (x - 3) \) are factors, their product \( (x + 2)(x - 3) = x^2 - x - 6 \) is also a factor. Dividing \( f(x) \) by \( x^2 - x - 6 \) yields \( (x + 1) \). Complete factorisation: \( x^3 - 7x - 6 = (x + 2)(x - 3)(x + 1) \).
In simple words: Find the unknown constants using the two factor zeros. Then use the product of the two known factors to divide the polynomial and find the third factor.

Exam Tip: Once you find the constants, the complete factorisation follows naturally - always verify by expanding the final factored form.

Question 22. (x - 2) is a factor of the expression \( x^3 + ax^2 + bx + 6 \). When this expression is divided by (x - 3), it leaves the remainder 3. Find the values of a and b.
Answer: Let \( f(x) = x^3 + ax^2 + bx + 6 \). Since \( (x - 2) \) is a factor, \( f(2) = 0 \): \( 8 + 4a + 2b + 6 = 0 \), giving \( 2a + b = -7 \), so \( b = -7 - 2a \) (Equation 1). By the remainder theorem, when \( f(x) \) is divided by \( (x - 3) \), the remainder is \( f(3) = 3 \): \( 27 + 9a + 3b + 6 = 3 \), giving \( 9a + 3b = -30 \), or \( 3a + b = -10 \). Substituting Equation 1: \( 3a + (-7 - 2a) = -10 \), which simplifies to \( a = -3 \). Then \( b = -7 - 2(-3) = -1 \).
In simple words: Use the factor condition to write one equation. Use the remainder theorem to write a second equation. Solve the system to find both unknowns.

Exam Tip: Remember the remainder theorem: when dividing by \( (x - c) \), the remainder is \( f(c) \). Always set up two separate equations when you have two pieces of information.

Question 23. If (x - 2) is a factor of the expression \( 2x^3 + ax^2 + bx - 14 \) and when the expression is divided by (x - 3), it leaves a remainder 52, find the values of a and b.
Answer: Let \( f(x) = 2x^3 + ax^2 + bx - 14 \). Since \( (x - 2) \) is a factor, \( f(2) = 0 \): \( 16 + 4a + 2b - 14 = 0 \), giving \( 2a + b = -1 \), so \( b = -1 - 2a \) (Equation 1). By the remainder theorem, \( f(3) = 52 \): \( 2(27) + a(9) + 3b - 14 = 52 \), which simplifies to \( 54 + 9a + 3b - 14 = 52 \), giving \( 9a + 3b = 12 \), or \( 3a + b = 4 \). Substituting Equation 1: \( 3a + (-1 - 2a) = 4 \), which gives \( a = 5 \). Then \( b = -1 - 2(5) = -11 \).
In simple words: Apply the factor theorem for the first condition and the remainder theorem for the second condition. Combine the two equations to solve for both unknowns.

Exam Tip: When dividing by a linear expression, the remainder is always a constant number. Use this systematically to create your second equation alongside the factor condition.

Question 24. If ax³ + 3x² + bx - 3 has a factor (2x + 3) and leaves remainder -3 when divided by (x + 2), find the values of a and b. With these values of a and b, factorise the given expression.
Answer: Let f(x) = ax³ + 3x² + bx - 3. Since (2x + 3) is a factor of f(x), by the factor theorem, f(-3/2) = 0. Substituting x = -3/2 and simplifying:
\[ a\left(-\frac{3}{2}\right)^3 + 3\left(-\frac{3}{2}\right)^2 + b\left(-\frac{3}{2}\right) - 3 = 0 \]
\[ -\frac{27a}{8} + \frac{27}{4} - \frac{3b}{2} - 3 = 0 \]
Multiplying through by 8 and simplifying gives: -27a - 12b + 30 = 0, or dividing by 3: -9a - 4b + 10 = 0, which gives 4b + 9a = 10 (Equation 1).
When f(x) is divided by (x + 2), the remainder is -3. By the remainder theorem, f(-2) = -3:
\[ a(-2)^3 + 3(-2)^2 + b(-2) - 3 = -3 \]
\[ -8a + 12 - 2b - 3 = -3 \]
\[ -8a - 2b = -12 \]
Dividing by -2: 4a + b = 6 (Equation 2). Multiplying Equation 2 by 4 gives 16a + 4b = 24. Subtracting Equation 1 from this: -7a = -14, so a = 2. Substituting back: b = 6 - 4(2) = -2.
With a = 2 and b = -2, we have f(x) = 2x³ + 3x² - 2x - 3. Factoring by grouping: f(x) = x²(2x + 3) - 1(2x + 3) = (x² - 1)(2x + 3) = (x - 1)(x + 1)(2x + 3).
In simple words: We use the factor theorem to set up two equations from the two given conditions. Solving these equations gives a = 2 and b = -2. With these values, the polynomial factors as (x - 1)(x + 1)(2x + 3).

Exam Tip: When using the factor and remainder theorems, carefully substitute the required values and simplify step by step. Always verify your final factorisation by expanding to check it matches the original polynomial.

Question 25. Given f(x) = ax² + bx + 2 and g(x) = bx² + ax + 1. If (x - 2) is a factor of f(x) but leaves the remainder -15 when it divides g(x), find the values of a and b. With these values of a and b, factorise the expression f(x) + g(x) + 4x² + 7x
Answer: Let f(x) = ax² + bx + 2 and g(x) = bx² + ax + 1. Since (x - 2) is a factor of f(x), by the factor theorem, f(2) = 0:
\[ a(2)^2 + b(2) + 2 = 0 \]
\[ 4a + 2b + 2 = 0 \]
Dividing by 2: 2a + b + 1 = 0, so b = -1 - 2a (Equation 1).
When g(x) is divided by (x - 2), the remainder is -15. By the remainder theorem, g(2) = -15:
\[ b(2)^2 + a(2) + 1 = -15 \]
\[ 4b + 2a + 1 = -15 \]
\[ 4b + 2a = -16 \]
Dividing by 2: 2b + a = -8. Substituting b = -1 - 2a from Equation 1: 2(-1 - 2a) + a = -8, which simplifies to -2 - 4a + a = -8, giving -3a = -6, so a = 2. Then b = -1 - 2(2) = -5.
With a = 2 and b = -5, we compute f(x) + g(x) + 4x² + 7x:
\[ (2x^2 - 5x + 2) + (-5x^2 + 2x + 1) + 4x^2 + 7x \]
\[ = 2x^2 - 5x^2 + 4x^2 - 5x + 2x + 7x + 2 + 1 \]
\[ = x^2 + 4x + 3 \]
\[ = x^2 + 3x + x + 3 \]
\[ = x(x + 3) + 1(x + 3) \]
\[ = (x + 1)(x + 3) \]
In simple words: We apply the factor and remainder theorems to find two equations. Solving gives a = 2 and b = -5. When we add the two functions with the extra terms and simplify, we get a quadratic that factors neatly as (x + 1)(x + 3).

Exam Tip: Carefully combine like terms when adding polynomials, and verify the final factorisation. A common error is dropping a term during simplification, so track the constant term and each power of x separately.

Multiple Choice Questions

Question 1. When 2x³ - x² - 3x + 5 is divided by 2x + 1, then the remainder is
(a) 6
(b) -6
(c) -3
(d) 0
Answer: (a) 6
In simple words: Use the remainder theorem: divide the polynomial by 2x + 1, which is the same as dividing by 2(x + 1/2). Substitute x = -1/2 into the polynomial to find the remainder directly, which equals 6.

Exam Tip: When dividing by a linear divisor (x - a), always use the remainder theorem to avoid long division—simply substitute the value of x that makes the divisor zero into the polynomial.

Question 2. If on dividing 4x² - 3kx + 5 by x + 2, the remainder is -3 then the value of k is
(a) 4
(b) -4
(c) 3
(d) -3
Answer: (b) -4
In simple words: Using the remainder theorem, substitute x = -2 into the polynomial 4x² - 3kx + 5 and set it equal to -3. This gives an equation you can solve to find k = -4.

Exam Tip: Always set the remainder equal to the given value and solve for the unknown. Double-check by substituting back into the equation to confirm your answer.

Question 3. If on dividing 2x³ + 6x² - (2k - 7)x + 5 by (x + 3), the remainder is k - 1 then the value of k is
(a) 2
(b) -2
(c) -3
(d) 3
Answer: (d) 3
In simple words: By the remainder theorem, substitute x = -3 into the polynomial and set it equal to k - 1. Simplifying: 2(-3)³ + 6(-3)² - (2k - 7)(-3) + 5 = k - 1, which becomes -54 + 54 + 6k - 21 + 5 = k - 1. Simplifying gives 6k - 16 = k - 1, so 5k = 15 and k = 3.

Exam Tip: When a remainder is given in terms of an unknown, set up the equation carefully using the remainder theorem and solve step by step, keeping track of all algebraic terms.

Question 4. If x + 1 is a factor of 3x³ + kx² + 7x + 4, then the value of k is
(a) -1
(b) 0
(c) 6
(d) 10
Answer: (c) 6
In simple words: If (x + 1) is a factor, then by the factor theorem, the polynomial equals zero when x = -1. Substitute x = -1: 3(-1)³ + k(-1)² + 7(-1) + 4 = 0, which gives -3 + k - 7 + 4 = 0. Solving: k - 6 = 0, so k = 6.

Exam Tip: For factor theorem questions, always set f(a) = 0 where (x - a) is the factor. Here, x + 1 means a = -1, so substitute carefully and solve the resulting equation.

Question 5. What must be subtracted from the polynomial x³ + x² - 2x + 1, so that the result is exactly divisible by (x - 3)?
(a) -31
(b) -30
(c) 30
(d) 31
Answer: (d) 31
In simple words: Let k be the number to subtract. The resulting polynomial is x³ + x² - 2x + 1 - k, and this must be divisible by (x - 3), so it must equal zero when x = 3. Substitute x = 3: 3³ + 3² - 2(3) + 1 - k = 0, which gives 27 + 9 - 6 + 1 - k = 0. Simplifying: 31 - k = 0, so k = 31.

Exam Tip: When asked what to subtract so a polynomial is divisible by a linear factor, find the remainder when dividing by that factor—that remainder is exactly what must be subtracted.

Question 6. A polynomial in 'x' is divided by (x - a) and for (x - a) to be a factor of this polynomial, the remainder should be
(a) -a
(b) 0
(c) a
(d) 2a
Answer: (b) 0
In simple words: By the factor theorem, (x - a) is a factor of a polynomial if and only if the remainder when dividing by (x - a) equals zero.

Exam Tip: This is a key definition: a polynomial has (x - a) as a factor if and only if substituting x = a into the polynomial gives zero.

Assertion-Reason Type Questions

Question. Assertion (A): The factorisation of ax² + bx + c is possible only if its discriminant = b² - 4ac < 0.
Reason (R): To factorise ax² + bx + c, split the coefficient of x into two real numbers such that their algebraic sum is b and their product is ac.
(1) Assertion (A) is true, but Reason (R) is false.
(2) Assertion (A) is false, but Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (2) Assertion (A) is false, but Reason (R) is true.
In simple words: The assertion wrongly states that factorisation needs a negative discriminant. In reality, factorisation over the real numbers happens when the discriminant is greater than or equal to zero (D ≥ 0). The reason correctly describes the middle-term splitting method, which is a valid technique for factoring quadratics when real roots exist.

Exam Tip: Remember that for real factorisation, the discriminant must be non-negative (D ≥ 0), not negative. The middle-term splitting method is the standard factorisation technique and works precisely when D ≥ 0.

Question. Given a polynomial f(x) = 2x³ - 7x² - 5x + 4.
Assertion (A): (x - 1) is not factor of f(x).
Reason (R): f(1) = -6.
(1) Assertion (A) is true, but Reason (R) is false.
(2) Assertion (A) is false, but Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
In simple words: Calculate f(1) = 2(1)³ - 7(1)² - 5(1) + 4 = 2 - 7 - 5 + 4 = -6. Since f(1) is not zero, by the factor theorem, (x - 1) is not a factor of f(x). The reason gives the exact value that confirms the assertion.

Exam Tip: Use the factor theorem: (x - a) is a factor if and only if f(a) = 0. If f(a) ≠ 0, then (x - a) is not a factor.

Question. Given f(x) = 16x³ - 8x² + 4x + 7.
Assertion (A): When we subtract 1 from f(x), the resulting polynomial is divisible by (2x + 1).
Reason (R): f(-1/2) = 1.
(1) Assertion (A) is true, but Reason (R) is false.
(2) Assertion (A) is false, but Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (3) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
In simple words: When we subtract 1, we get g(x) = 16x³ - 8x² + 4x + 6. For (2x + 1) to be a factor, g(-1/2) must equal zero. Compute: g(-1/2) = 16(-1/8) - 8(1/4) + 4(-1/2) + 6 = -2 - 2 - 2 + 6 = 0. Also, f(-1/2) = -2 - 2 - 2 + 7 = 1. Since subtracting 1 from f gives a polynomial with f(-1/2) = 1 as the remainder, and that remainder is exactly 1, the new polynomial g has g(-1/2) = 0, confirming divisibility.

Exam Tip: When checking divisibility by a factor, use the factor theorem: (2x + 1) divides g(x) if and only if g(-1/2) = 0. Notice that f(-1/2) = 1 tells you that the remainder when dividing f by (2x + 1) is 1, so subtracting 1 removes this remainder.

Chapter Test

Question 1. Find the remainder when 2x³ - 3x² + 4x + 7 is divided by
(i) x - 2
(ii) x + 3
(iii) 2x + 1
Answer: (i) By the remainder theorem, when f(x) = 2x³ - 3x² + 4x + 7 is divided by (x - 2), the remainder is f(2). Computing: f(2) = 2(2)³ - 3(2)² + 4(2) + 7 = 16 - 12 + 8 + 7 = 19. Hence, the remainder is 19.
(ii) When f(x) is divided by (x + 3) or (x - (-3)), the remainder is f(-3). Computing: f(-3) = 2(-3)³ - 3(-3)² + 4(-3) + 7 = -54 - 27 - 12 + 7 = -86. Hence, the remainder is -86.
(iii) When f(x) is divided by (2x + 1) or 2(x - (-1/2)), the remainder is f(-1/2). Computing: f(-1/2) = 2(-1/8) - 3(1/4) + 4(-1/2) + 7 = -1/4 - 3/4 - 2 + 7 = 16/4 = 4. Hence, the remainder is 4.
In simple words: Plug in the value of x that makes each divisor zero into the polynomial to get the remainder quickly—this is much faster than doing long division.

Exam Tip: For each part, identify the correct value to substitute. For (x - a), use x = a; for (x + a), use x = -a; for (2x + a), use x = -a/2.

Question 2. When 2x³ - 9x² + 10x - p is divided by (x + 1), the remainder is -24. Find the value of p.
Answer: By the remainder theorem, when f(x) = 2x³ - 9x² + 10x - p is divided by (x + 1) or (x - (-1)), the remainder is f(-1). Given that the remainder is -24:
\[ f(-1) = 2(-1)^3 - 9(-1)^2 + 10(-1) - p = -24 \]
\[ = -2 - 9 - 10 - p = -24 \]
\[ = -21 - p = -24 \]
Solving: p = 24 - 21 = 3. Hence, p = 3.
In simple words: Substitute x = -1 into the polynomial and set it equal to the given remainder. This gives you a simple equation to solve for the unknown coefficient p.

Exam Tip: Always set up the equation by substituting the test value into the polynomial and equating it to the given remainder. Clear arithmetic and careful handling of signs are essential.

Question 3. If (2x - 3) is a factor of 6x² + x + a, find the value of a. With this value of a, factorise the given expression.
Answer: By the factor theorem, if (2x - 3) is a factor of f(x) = 6x² + x + a, then f(3/2) = 0. Substituting x = 3/2:
\[ 6\left(\frac{3}{2}\right)^2 + \frac{3}{2} + a = 0 \]
\[ 6 \cdot \frac{9}{4} + \frac{3}{2} + a = 0 \]
\[ \frac{27}{2} + \frac{3}{2} + a = 0 \]
\[ \frac{30}{2} + a = 0 \]
\[ 15 + a = 0 \]
So a = -15. With a = -15, the polynomial becomes f(x) = 6x² + x - 15. To factorise, we look for two numbers that multiply to 6 × (-15) = -90 and add to 1. These are 10 and -9. Rewriting:
\[ f(x) = 6x^2 + 10x - 9x - 15 \]
\[ = 2x(3x + 5) - 3(3x + 5) \]
\[ = (2x - 3)(3x + 5) \]
Hence, a = -15 and the factorisation is 6x² + x - 15 = (2x - 3)(3x + 5).
In simple words: Use the factor theorem to find the unknown coefficient by setting the polynomial to zero at the factor's root. Then use the middle-term splitting method to factor the resulting quadratic expression.

Exam Tip: When finding an unknown coefficient, always use the factor theorem carefully. After finding the coefficient, verify your factorisation by expanding the factors to confirm they match the original polynomial.

Question 4. When \( 3x^2 - 5x + p \) is divided by (x - 2), the remainder is 3. Find the value of p. Also factorise the polynomial \( 3x^2 - 5x + p - 3 \).
Answer: Using the remainder theorem, when you divide f(x) by (x - a), what is left over is f(a). Since f(x) = \( 3x^2 - 5x + p \) and we divide by (x - 2), the remainder equals f(2).

We know the remainder is 3, so f(2) = 3.

\[ \Rightarrow 3(2)^2 - 5(2) + p = 3 \]
\[ \Rightarrow 12 - 10 + p = 3 \]
\[ \Rightarrow p + 2 = 3 \]
\[ \Rightarrow p = 1 \]

Now substitute p = 1 into \( 3x^2 - 5x + p - 3 \):
\[ 3x^2 - 5x + 1 - 3 = 3x^2 - 5x - 2 \]
\[ = 3x^2 - 6x + x - 2 \]
\[ = 3x(x - 2) + 1(x - 2) \]
\[ = (3x + 1)(x - 2) \]

Therefore, p = 1 and the factors are (3x + 1) and (x - 2).
In simple words: Put x = 2 into the expression. This helps you find p. Once you know p, you can break down the new expression into two simpler parts that multiply together.

Exam Tip: Always apply the remainder theorem correctly - substitute the value that makes the divisor zero. Verify your answer by expanding the factors to check they match the original polynomial.

Question 5. Prove that (5x + 4) is a factor of \( 5x^3 + 4x^2 - 5x - 4 \). Hence, factorise the given polynomial completely.
Answer: By the factor theorem, (x - b) is a factor of f(x) if f(b) = 0.

Let f(x) = \( 5x^3 + 4x^2 - 5x - 4 \)

Since (5x + 4) or \( 5(x - (-\frac{4}{5})) \) is proposed as a factor, we find f(-4/5):

\[ f(-\frac{4}{5}) = 5(-\frac{4}{5})^3 + 4(-\frac{4}{5})^2 - 5(-\frac{4}{5}) - 4 \]
\[ = 5(-\frac{64}{125}) + 4(\frac{16}{25}) + 4 - 4 \]
\[ = -\frac{64}{25} + \frac{64}{25} \]
\[ = 0 \]

Since f(-4/5) = 0, (5x + 4) is indeed a factor of f(x).

Now we factor the polynomial:
\[ 5x^3 + 4x^2 - 5x - 4 = x^2(5x + 4) - 1(5x + 4) \]
\[ = (x^2 - 1)(5x + 4) \]
\[ = (x^2 - 1^2)(5x + 4) \]
\[ = (x - 1)(x + 1)(5x + 4) \]
In simple words: Check if (5x + 4) makes the expression equal zero. If it does, then (5x + 4) is definitely a factor. Then group the terms to pull out this factor and break down what remains further.

Exam Tip: When checking divisibility by a linear factor like (5x + 4), solve 5x + 4 = 0 to find x = -4/5, then substitute this value. Group terms strategically to extract common factors before using difference of squares.

Question 6. Use factor theorem to factorise the following polynomials completely:
(i) \( 4x^3 + 4x^2 - 9x - 9 \)
(ii) \( x^3 - 19x - 30 \)
(iii) \( 2x^3 - x^2 - 13x - 6 \)
Answer:
(i) Let f(x) = \( 4x^3 + 4x^2 - 9x - 9 \)

Test x = -1:
\[ f(-1) = 4(-1)^3 + 4(-1)^2 - 9(-1) - 9 = -4 + 4 + 9 - 9 = 0 \]

Since f(-1) = 0, (x + 1) is a factor. Dividing the polynomial by (x + 1):
\[ 4x^3 + 4x^2 - 9x - 9 = (x + 1)(4x^2 - 9) \]
\[ = (x + 1)((2x)^2 - 3^2) \]
\[ = (x + 1)(2x - 3)(2x + 3) \]

(ii) Let f(x) = \( x^3 - 19x - 30 \)

Test x = -2:
\[ f(-2) = (-2)^3 - 19(-2) - 30 = -8 + 38 - 30 = 0 \]

Since f(-2) = 0, (x + 2) is a factor. Dividing by (x + 2):
\[ x^3 - 19x - 30 = (x + 2)(x^2 - 2x - 15) \]
\[ = (x + 2)(x - 5x + 3x - 15) \]
\[ = (x + 2)(x(x - 5) + 3(x - 5)) \]
\[ = (x + 2)(x + 3)(x - 5) \]

(iii) Let f(x) = \( 2x^3 - x^2 - 13x - 6 \)

Test x = -2:
\[ f(-2) = 2(-2)^3 - (-2)^2 - 13(-2) - 6 = -16 - 4 + 26 - 6 = 0 \]

Since f(-2) = 0, (x + 2) is a factor. Dividing by (x + 2):
\[ 2x^3 - x^2 - 13x - 6 = (x + 2)(2x^2 - 5x - 3) \]
\[ = (x + 2)(2x^2 - 6x + x - 3) \]
\[ = (x + 2)(2x(x - 3) + 1(x - 3)) \]
\[ = (x + 2)(2x + 1)(x - 3) \]
In simple words: For each polynomial, try small whole numbers and their negatives until the expression becomes zero. Once you find one, that tells you a factor. Then divide to get a simpler expression and break it down further.

Exam Tip: Use the rational root theorem - test factors of the constant term. Write out all working carefully, showing the division process. Always verify your final factors by multiplying them back together.

Question 7. If \( x^3 - 2x^2 + px + q \) has a factor (x + 2) and leaves a remainder 9 when divided by (x + 1), find the values of p and q. With these values of p and q, factorise the given polynomial completely.
Answer: By the factor theorem, (x - b) is a factor of f(x) if f(b) = 0.

Let f(x) = \( x^3 - 2x^2 + px + q \)

Given that (x + 2) is a factor, so f(-2) = 0:
\[ (-2)^3 - 2(-2)^2 + p(-2) + q = 0 \]
\[ -8 - 8 - 2p + q = 0 \]
\[ -2p + q = 16 \]
\[ q = 2p + 16 \quad \text{(Equation 1)} \]

By the remainder theorem, when divided by (x + 1), the remainder is f(-1) = 9:
\[ (-1)^3 - 2(-1)^2 + p(-1) + q = 9 \]
\[ -1 - 2 - p + q = 9 \]
\[ -3 - p + q = 9 \]
\[ q - p = 12 \]

Substitute q = 2p + 16 from Equation 1:
\[ 2p + 16 - p = 12 \]
\[ p + 16 = 12 \]
\[ p = -4 \]

Then q = 2(-4) + 16 = -8 + 16 = 8

So f(x) = \( x^3 - 2x^2 - 4x + 8 \)

Since (x + 2) is a factor, divide f(x) by (x + 2):
\[ x^3 - 2x^2 - 4x + 8 = (x + 2)(x^2 - 4x + 4) \]
\[ = (x + 2)(x^2 - 2 \times 2 \times x + 2^2) \]
\[ = (x + 2)(x - 2)^2 \]
In simple words: Use the two conditions to write down two equations. Solve them together to find p and q. Then divide the polynomial using the known factor to get the final answer.

Exam Tip: Apply both conditions (factor and remainder) to set up two simultaneous equations. Recognize perfect square trinomials like \( x^2 - 4x + 4 = (x-2)^2 \) to complete the factorization.

Question 8. If (x + 3) and (x - 4) are factors of \( x^3 + ax^2 - bx + 24 \), find the values of a and b. With these values of a and b, factorise the given expression.
Answer: By the factor theorem, (x - c) is a factor of f(x) if f(c) = 0.

Let f(x) = \( x^3 + ax^2 - bx + 24 \)

Given (x + 3) is a factor, so f(-3) = 0:
\[ (-3)^3 + a(-3)^2 - b(-3) + 24 = 0 \]
\[ -27 + 9a + 3b + 24 = 0 \]
\[ 9a + 3b - 3 = 0 \]
\[ 9a + 3b = 3 \]
\[ 3a + b = 1 \]
\[ b = 1 - 3a \quad \text{(Equation 1)} \]

Given (x - 4) is a factor, so f(4) = 0:
\[ (4)^3 + a(4)^2 - b(4) + 24 = 0 \]
\[ 64 + 16a - 4b + 24 = 0 \]
\[ 16a - 4b + 88 = 0 \]
\[ 16a - 4b = -88 \]
\[ 4a - b = -22 \]

Substitute b = 1 - 3a from Equation 1:
\[ 4a - (1 - 3a) = -22 \]
\[ 4a - 1 + 3a = -22 \]
\[ 7a - 1 = -22 \]
\[ 7a = -21 \]
\[ a = -3 \]

Then b = 1 - 3(-3) = 1 + 9 = 10

So f(x) = \( x^3 - 3x^2 - 10x + 24 \)

Since (x + 3) and (x - 4) are factors, their product (x + 3)(x - 4) = \( x^2 - x - 12 \) is also a factor. Dividing:
\[ x^3 - 3x^2 - 10x + 24 = (x - 2)(x^2 - x - 12) \]
\[ = (x - 2)(x - 4x + 3x - 12) \]
\[ = (x - 2)(x(x - 4) + 3(x - 4)) \]
\[ = (x - 2)(x + 3)(x - 4) \]
In simple words: Both given factors make the polynomial zero. This gives you two equations. Solve them to find a and b. Then you know the polynomial, and you can break it down using division.

Exam Tip: When given two factors, use both to create two equations immediately. The product of both factors will divide the cubic evenly, leaving a linear quotient.

Question 9. If (2x + 1) is a factor of both the expressions \( 2x^2 - 5x + p \) and \( 2x^2 + 5x + q \), find the values of p and q. Hence, find the other factors of both the polynomials.
Answer: By the factor theorem, (x - b) is a factor of f(x) if f(b) = 0.

Let f(x) = \( 2x^2 - 5x + p \)

Since (2x + 1) or \( 2(x - (-\frac{1}{2})) \) is a factor, we have f(-1/2) = 0:
\[ 2(-\frac{1}{2})^2 - 5(-\frac{1}{2}) + p = 0 \]
\[ 2(\frac{1}{4}) + \frac{5}{2} + p = 0 \]
\[ \frac{1}{2} + \frac{5}{2} + p = 0 \]
\[ \frac{6}{2} + p = 0 \]
\[ 3 + p = 0 \]
\[ p = -3 \]

So f(x) = \( 2x^2 - 5x - 3 \)
\[ = 2x^2 - 6x + x - 3 \]
\[ = 2x(x - 3) + 1(x - 3) \]
\[ = (2x + 1)(x - 3) \]

Let g(x) = \( 2x^2 + 5x + q \)

Since (2x + 1) is a factor, g(-1/2) = 0:
\[ 2(-\frac{1}{2})^2 + 5(-\frac{1}{2}) + q = 0 \]
\[ 2(\frac{1}{4}) - \frac{5}{2} + q = 0 \]
\[ \frac{1}{2} - \frac{5}{2} + q = 0 \]
\[ -\frac{4}{2} + q = 0 \]
\[ -2 + q = 0 \]
\[ q = 2 \]

So g(x) = \( 2x^2 + 5x + 2 \)
\[ = 2x^2 + 4x + x + 2 \]
\[ = 2x(x + 2) + 1(x + 2) \]
\[ = (2x + 1)(x + 2) \]
In simple words: A factor means plugging in a value makes the expression zero. Find that value from 2x + 1 = 0. Put it in both expressions to find p and q. Then break each expression into its factors.

Exam Tip: Solve the linear factor equation to get the substitution value. The other factor will always be linear in a quadratic. Check your answer by expanding the two factors.

Question 10. If a polynomial f(x) = \( x^4 - 2x^3 + 3x^2 - ax - b \) leaves remainders 5 and 19 when divided by (x - 1) and (x + 1) respectively, find the values of a and b. Hence, determine the remainder when f(x) is divided by (x - 2).
Answer: By the remainder theorem, when you divide f(x) by (x - a), the remainder that results is f(a).

Let f(x) = \( x^4 - 2x^3 + 3x^2 - ax - b \)

When divided by (x + 1), the remainder is f(-1) = 19:
\[ (-1)^4 - 2(-1)^3 + 3(-1)^2 - a(-1) - b = 19 \]
\[ 1 + 2 + 3 + a - b = 19 \]
\[ a - b + 6 = 19 \]
\[ a - b = 13 \]
\[ a = 13 + b \quad \text{(Equation 1)} \]

When divided by (x - 1), the remainder is f(1) = 5:
\[ (1)^4 - 2(1)^3 + 3(1)^2 - a(1) - b = 5 \]
\[ 1 - 2 + 3 - a - b = 5 \]
\[ 2 - a - b = 5 \]
\[ -a - b = 3 \]

Substitute a = 13 + b from Equation 1:
\[ -(13 + b) - b = 3 \]
\[ -13 - b - b = 3 \]
\[ -13 - 2b = 3 \]
\[ -2b = 16 \]
\[ b = -8 \]

Then a = 13 + (-8) = 5

So f(x) = \( x^4 - 2x^3 + 3x^2 - 5x + 8 \)

When divided by (x - 2), the remainder is f(2):
\[ f(2) = (2)^4 - 2(2)^3 + 3(2)^2 - 5(2) + 8 \]
\[ = 16 - 16 + 12 - 10 + 8 \]
\[ = 10 \]
In simple words: Each remainder condition gives you one equation. When you know a and b, you can find the remainder for any divisor by just substituting that divisor's value.

Exam Tip: Set up two equations from the two remainder conditions. Solve the system carefully. Always verify by substituting back into both original remainder conditions.

Question 11. When a polynomial f(x) is divided by (x - 1), the remainder is 5 and when it is divided by (x - 2), the remainder is 7. Find the remainder when it is divided by (x - 1)(x - 2).
Answer: By the remainder theorem, when f(x) is divided by (x - a), the remainder left is f(a).

Given that when f(x) is divided by (x - 1), the remainder is 5:
\[ f(1) = 5 \]

Given that when f(x) is divided by (x - 2), the remainder is 7:
\[ f(2) = 7 \]

Now suppose when f(x) is divided by (x - 1)(x - 2), we get:
Quotient = q(x)
Remainder = ax + b

So: f(x) = (x - 1)(x - 2)q(x) + ax + b

Substituting x = 1:
\[ f(1) = (1 - 1)(1 - 2)q(1) + a(1) + b = 5 \]
\[ 0 + a + b = 5 \]
\[ a + b = 5 \]
\[ a = 5 - b \quad \text{(Equation 1)} \]

Substituting x = 2:
\[ f(2) = (2 - 1)(2 - 2)q(2) + a(2) + b = 7 \]
\[ 0 + 2a + b = 7 \]
\[ 2a + b = 7 \]

Substitute a = 5 - b from Equation 1:
\[ 2(5 - b) + b = 7 \]
\[ 10 - 2b + b = 7 \]
\[ 10 - b = 7 \]
\[ b = 3 \]

Then a = 5 - 3 = 2

Therefore, the remainder is ax + b = 2x + 3
In simple words: When dividing by a product of two linear factors, the remainder must be linear (ax + b). Use the two remainder values to find a and b.

Exam Tip: The degree of the remainder is always one less than the degree of the divisor. For a quadratic divisor, the remainder must be linear. Substitute the roots of the divisor into the division formula to set up equations.

Question 12. The polynomial \( 3x^3 + 8x^2 - 15x + k \) has (x - 1) as a factor. Find the value of k. Hence factorise the resulting polynomial completely.
Answer: If (x - 1) is a factor, then setting x = 1 gives a remainder of zero. Substituting x = 1 into \( 3x^3 + 8x^2 - 15x + k \):

\[ 3(1)^3 + 8(1)^2 - 15(1) + k = 0 \]
\[ 3 + 8 - 15 + k = 0 \]
\[ 11 - 15 + k = 0 \]
\[ k - 4 = 0 \]
\[ k = 4 \]

So the polynomial is \( 3x^3 + 8x^2 - 15x + 4 \)

Now divide this polynomial by (x - 1):
\[ 3x^3 + 8x^2 - 15x + 4 = (x - 1)(3x^2 + 11x - 4) \]

Next, factor \( 3x^2 + 11x - 4 \):
\[ 3x^2 + 11x - 4 = 3x^2 + 12x - x - 4 \]
\[ = 3x(x + 4) - 1(x + 4) \]
\[ = (3x - 1)(x + 4) \]

Therefore:
\[ 3x^3 + 8x^2 - 15x + 4 = (x - 1)(3x - 1)(x + 4) \]
In simple words: If (x - 1) is a factor, then putting x = 1 must make the whole expression zero. Use this to find k. Then divide by (x - 1) and break down what is left.

Exam Tip: Always check your value of k by substituting back. When factoring the resulting quadratic, look for two numbers that multiply to give ac and add to give b.

Question 13. While factorising a given polynomial, using remainder and factor theorem, a student finds that (2x + 1) is a factor of \( 2x^3 + 7x^2 + 2x - 3 \). (a) Is the student's solution correct stating that (2x + 1) is a factor of the given polynomial? Give a valid reason for your answer. (b) Factorise the given polynomial completely.
Answer: (a) If (2x + 1) is a factor, then setting 2x + 1 = 0 gives x = -1/2. Substituting into \( 2x^3 + 7x^2 + 2x - 3 \):

\[ 2(-\frac{1}{2})^3 + 7(-\frac{1}{2})^2 + 2(-\frac{1}{2}) - 3 \]
\[ = 2(-\frac{1}{8}) + 7(\frac{1}{4}) - 1 - 3 \]
\[ = -\frac{1}{4} + \frac{7}{4} - 4 \]
\[ = \frac{6}{4} - 4 \]
\[ = \frac{3}{2} - 4 \]
\[ \neq 0 \]

Since the result is not zero, the student's solution is incorrect. (2x + 1) is not a factor of the given polynomial.

(b) Let f(x) = \( 2x^3 + 7x^2 + 2x - 3 \)

Test x = -3:
\[ f(-3) = 2(-3)^3 + 7(-3)^2 + 2(-3) - 3 = -54 + 63 - 6 - 3 = 0 \]

So (x + 3) is a factor. Dividing by (x + 3):
\[ 2x^3 + 7x^2 + 2x - 3 = (x + 3)(2x^2 + x - 1) \]

Now factor \( 2x^2 + x - 1 \):
\[ 2x^2 + x - 1 = 2x^2 + 2x - x - 1 \]
\[ = 2x(x + 1) - 1(x + 1) \]
\[ = (2x - 1)(x + 1) \]

Therefore:
\[ 2x^3 + 7x^2 + 2x - 3 = (x + 3)(2x - 1)(x + 1) \]
In simple words: To check if something is a factor, substitute the value that makes it zero. If the whole expression becomes zero too, then it is a factor. If not, it is not a factor. The correct factor must work.

Exam Tip: Always verify claimed factors by substitution before accepting them. When the given factor is incorrect, systematically test other possible factors using small integers. Show clear reasoning for why a factor does or does not work.

Question 1. Show that (2x + 1) is not a factor of 2x³ + 7x² + 2x - 3.
Answer: To check if (2x + 1) is a factor, we use the Factor Theorem. If (2x + 1) is a factor, then the polynomial should equal zero when we set 2x + 1 = 0, which gives x = -1/2.

Substituting x = -1/2 into 2x³ + 7x² + 2x - 3:

\[ 2 \times \left( -\frac{1}{2} \right)^3 + 7 \times \left( -\frac{1}{2} \right)^2 + 2 \times \left( -\frac{1}{2} \right) - 3 \]

\[ = 2 \times \left( -\frac{1}{8} \right) + 7 \times \frac{1}{4} - 1 - 3 \]

\[ = -\frac{1}{4} + \frac{7}{4} - 4 \]

\[ = \frac{-1 + 7}{4} - 4 \]

\[ = \frac{6}{4} - 4 \]

\[ = \frac{6 - 16}{4} \]

\[ = -\frac{10}{4} = -\frac{5}{2} \]

Since the remainder is -5/2, which is not equal to zero, (2x + 1) is not a factor of the given polynomial.
In simple words: We plug in the value that makes 2x + 1 equal to zero. If we get zero as the final answer, then (2x + 1) is a factor. Since we got -5/2 instead, (2x + 1) is not a factor.

Exam Tip: Always remember the Factor Theorem - a linear expression (ax + b) is a factor if and only if the polynomial equals zero when you substitute the value that makes (ax + b) = 0.

Question 2. Show that x - 1/2 is a factor of 2x³ + 7x² + 2x - 3.
Answer: To verify that x - 1/2 is a factor, we substitute x = 1/2 into the polynomial 2x³ + 7x² + 2x - 3.

\[ 2 \times \left( \frac{1}{2} \right)^3 + 7 \times \left( \frac{1}{2} \right)^2 + 2 \times \frac{1}{2} - 3 \]

\[ = 2 \times \frac{1}{8} + 7 \times \frac{1}{4} + 1 - 3 \]

\[ = \frac{1}{4} + \frac{7}{4} - 2 \]

\[ = \frac{8}{4} - 2 \]

\[ = 2 - 2 = 0 \]

Since the remainder equals zero, x - 1/2 is a factor of the polynomial. This also means 2x - 1 is a factor (multiply the expression by 2).
In simple words: When we replace x with 1/2, the whole expression becomes zero. This proves that x - 1/2 is a factor.

Exam Tip: When the polynomial gives zero after substitution, always state this clearly - "remainder equals zero" - as this is the key phrase that shows you understand the Factor Theorem.

Question 3. Divide 2x³ + 7x² + 2x - 3 by 2x - 1 and factorize the polynomial completely.
Answer: Using polynomial long division to divide 2x³ + 7x² + 2x - 3 by 2x - 1:

The quotient obtained is x² + 4x + 3. We now need to factorize this quadratic expression.

\[ x^2 + 4x + 3 = x^2 + 3x + x + 3 \]

\[ = x(x + 3) + 1(x + 3) \]

\[ = (x + 1)(x + 3) \]

Therefore, the complete factorization of 2x³ + 7x² + 2x - 3 is:

\[ 2x^3 + 7x^2 + 2x - 3 = (2x - 1)(x + 1)(x + 3) \]
In simple words: We first divide the big polynomial by (2x - 1) to get a smaller polynomial. Then we break down that smaller polynomial into two simple factors by grouping.

Exam Tip: Always verify your complete factorization by multiplying all three factors back together - if you get the original polynomial, your answer is correct. Show the grouping step clearly in the quadratic factorization.