CBSE Class 10 Mathematics Polynomials Assignment Set 02

Important Concepts

Take a Look

1. Polynomials: An algebraic expression of the form \( a_0 x^n + a_1 x^{n-1} + a_2 x^{n-2} + \dots + a_{n-1} x + a_n \), where the coefficients \( a_0, a_1, a_2, \dots, a_n \) are real constants and the power \( n \) is a non-negative integer, is defined as a polynomial in the variable \( x \).

2. Zeroes of a Polynomial: A real value \( k \) is called a zero of the polynomial \( p(x) \) if and only if substituting \( k \) in place of \( x \) results in \( p(k) = 0 \).

3. Graph of Polynomial:
i. The graphical representation of a linear polynomial \( ax+b \) is always a straight line.
ii. The graph of a quadratic polynomial \( p(x) = ax^2 + bx + c \) is a parabola that opens upwards (like a U-shape) when \( a > 0 \).
iii. The graph of a quadratic polynomial \( p(x) = ax^2 + bx + c \) is a parabola that opens downwards (like an inverted U-shape) when \( a < 0 \).

Discriminant of a quadratic polynomial:
For the quadratic polynomial \( p(x) = ax^2 + bx + c \), the expression \( b^2 - 4ac \) is defined as its discriminant, denoted by the letter 'D'.

\( \implies D = b^2 - 4ac \)

i. If \( D > 0 \), the parabola representing \( p(x) \) will cross the x-axis at two distinct locations. The x-coordinates of these points of intersection are the real zeroes of \( p(x) \).
ii. If \( D = 0 \), the parabola representing \( p(x) \) will touch the x-axis at precisely one point.

\( \implies p(x) \text{ will have only one distinct real zero.} \)
iii. If \( D < 0 \), the parabola representing \( p(x) \) will neither cross nor touch the x-axis.

\( \implies p(x) \text{ does not possess any real zeroes.} \)

4. Relationship between the zeroes and the coefficients of a polynomial:
i. If \( \alpha \) and \( \beta \) are the zeroes of the quadratic polynomial \( p(x) = ax^2 + bx + c \), then:
Sum of zeroes = \( \alpha + \beta = -\frac{b}{a} = \frac{-\text{Coefficient of } x}{\text{Coefficient of } x^2} \)
Product of zeroes = \( \alpha\beta = \frac{c}{a} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} \)

ii. If \( \alpha \), \( \beta \), and \( \gamma \) are the zeroes of the cubic polynomial \( p(x) = ax^3 + bx^2 + cx + d \), then:
Sum of zeroes = \( \alpha + \beta + \gamma = -\frac{b}{a} = \frac{-\text{Coefficient of } x^2}{\text{Coefficient of } x^3} \)
Sum of the products taken two at a time = \( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{\text{Coefficient of } x}{\text{Coefficient of } x^3} \)
Product of zeroes = \( \alpha\beta\gamma = -\frac{d}{a} = \frac{-\text{Constant term}}{\text{Coefficient of } x^3} \)

iii. If \( \alpha \) and \( \beta \) are the zeroes of a quadratic polynomial \( p(x) \), then the polynomial is given by:
\( p(x) = x^2 - (\alpha + \beta)x + \alpha\beta \)

iv. If \( \alpha \), \( \beta \), and \( \gamma \) are the zeroes of a cubic polynomial \( p(x) \), then the polynomial is given by:
\( p(x) = x^3 - (\alpha + \beta + \gamma)x^2 + (\alpha\beta + \beta\gamma + \gamma\alpha)x - \alpha\beta\gamma \)

5. Division Algorithm for polynomials:
If \( f(x) \) and \( g(x) \) are any two polynomials with \( g(x) \neq 0 \), we can find unique polynomials \( q(x) \) and \( r(x) \) such that:
\( f(x) = q(x) \cdot g(x) + r(x) \)
where either \( r(x) = 0 \) or the degree of \( r(x) \) is strictly less than the degree of \( g(x) \).
If the remainder becomes zero or its degree falls below the degree of the divisor, the division process cannot be continued further.
 

Level-I
 

Question. The zeroes of the polynomial 2x2-3x-2 are
(a) 1, 2
(b) -1/2, 1
(c) 1/2, -2
(d) -1/2, 2
Answer: (d) -1/2, 2
To find the zeroes, we set the quadratic expression to zero and factor it:
\( 2x^2 - 3x - 2 = 0 \)
\( 2x^2 - 4x + x - 2 = 0 \)
\( 2x(x - 2) + 1(x - 2) = 0 \)
\( (2x + 1)(x - 2) = 0 \)
This gives us:
\( 2x + 1 = 0 \implies x = -\frac{1}{2} \)
\( x - 2 = 0 \implies x = 2 \)
Therefore, the zeroes are \( -\frac{1}{2} \) and \( 2 \).
In simple words: Setting the quadratic polynomial equal to zero and factoring by splitting the middle term gives the roots \( -1/2 \) and \( 2 \).

Exam Tip: Factoring quadratic polynomials is the fastest method to find zeroes. Always double-check your sign signs when factoring.

Question. If \(\alpha\) and \(\beta\) are zeroes of the polynomial 2x2+7x-3, then the value of \(\alpha^2 + \beta^2\) is
(a) 49/4
(b) 37/4
(c) 61/4
(d) 61/2
Answer: (c) 61/4
For the polynomial \( 2x^2 + 7x - 3 \), we identify the coefficients: \( a = 2, b = 7, c = -3 \).
The sum of zeroes is:
\( \alpha + \beta = -\frac{b}{a} = -\frac{7}{2} \)
The product of zeroes is:
\( \alpha\beta = \frac{c}{a} = -\frac{3}{2} \)
Using the algebraic identity:
\( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \)
Substituting our values:
\( \alpha^2 + \beta^2 = \left(-\frac{7}{2}\right)^2 - 2\left(-\frac{3}{2}\right) \)
\( \alpha^2 + \beta^2 = \frac{49}{4} + 3 \)
\( \alpha^2 + \beta^2 = \frac{49 + 12}{4} = \frac{61}{4} \)
In simple words: First calculate the sum of zeroes as \( -7/2 \) and the product of zeroes as \( -3/2 \). Use these values in the formula \( (\alpha+\beta)^2 - 2\alpha\beta \) to find the result.

Exam Tip: Memorize the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \) as it is frequently tested in quadratic polynomial questions.

Question. If the polynomial 6x3+16x2+px -5 is exactly divisible by 3x+5 , then the value of p is
(a) -7
(b) -5
(c) 5
(d) 7
Answer: (d) 7
Let the polynomial be \( f(x) = 6x^3 + 16x^2 + px - 5 \). Since it is exactly divisible by \( 3x + 5 \), by the Remainder Theorem, \( f\left(-\frac{5}{3}\right) = 0 \).
Substituting \( x = -\frac{5}{3} \):
\( 6\left(-\frac{5}{3}\right)^3 + 16\left(-\frac{5}{3}\right)^2 + p\left(-\frac{5}{3}\right) - 5 = 0 \)
\( 6\left(-\frac{125}{27}\right) + 16\left(\frac{25}{9}\right) - \frac{5p}{3} - 5 = 0 \)
\( 2\left(-\frac{125}{9}\right) + \frac{400}{9} - \frac{5p}{3} - 5 = 0 \)
\( -\frac{250}{9} + \frac{400}{9} - \frac{5p}{3} - 5 = 0 \)
\( \frac{150}{9} - 5 - \frac{5p}{3} = 0 \)
\( \frac{50}{3} - 5 - \frac{5p}{3} = 0 \)
\( \frac{50 - 15}{3} = \frac{5p}{3} \)
\( \frac{35}{3} = \frac{5p}{3} \)
\( 5p = 35 \)

\( \implies p = 7 \)
In simple words: Since \( 3x + 5 \) divides the polynomial cleanly, putting \( x = -5/3 \) into the polynomial must give a remainder of zero. Solving this equation yields \( p = 7 \).

Exam Tip: Substituting the root of the linear divisor into the polynomial is much faster and less error-prone than using polynomial long division.

Question. If 2 is a zero of both the polynomials 3x2+ax-14 and 2x3+bx2+x-2, then the value of a-2b is
(a) -1
(b) 5
(c) 9
(d) -9
Answer: (c) 9
Let \( p(x) = 3x^2 + ax - 14 \). Since 2 is a zero of \( p(x) \), we have \( p(2) = 0 \):
\( 3(2)^2 + a(2) - 14 = 0 \)
\( 12 + 2a - 14 = 0 \)
\( 2a - 2 = 0 \)

\( \implies a = 1 \)

Let \( q(x) = 2x^3 + bx^2 + x - 2 \). Since 2 is also a zero of \( q(x) \), we have \( q(2) = 0 \):
\( 2(2)^3 + b(2)^2 + 2 - 2 = 0 \)
\( 16 + 4b = 0 \)
\( 4b = -16 \)

\( \implies b = -4 \)

Now, we calculate the value of \( a - 2b \):
\( a - 2b = 1 - 2(-4) = 1 + 8 = 9 \).
In simple words: Plug \( x = 2 \) into both equations. This gives \( a = 1 \) and \( b = -4 \). Substituting these into \( a - 2b \) results in \( 9 \).

Exam Tip: Solve for each variable separately first. Be extremely careful with signs when performing the final subtraction, especially with a negative value for \( b \).

Question. A quadratic polynomial whose product and sum of zeroes are 1/3 and \(\sqrt{2}\) respectively is
(a) \(3x^2 - x + 3\sqrt{2}\)
(b) \(3x^2 + x - 3\sqrt{2}\)
(c) \(3x^2 + 3\sqrt{2}x + 1\)
(d) \(3x^2 - 3\sqrt{2}x + 1\)
Answer: (d) \(3x^2 - 3\sqrt{2}x + 1\)
The general formula for a quadratic polynomial in terms of the sum \( S \) and product \( P \) of its zeroes is:
\( k(x^2 - Sx + P) \)
Given \( S = \sqrt{2} \) and \( P = \frac{1}{3} \), we can write:
\( k\left(x^2 - \sqrt{2}x + \frac{1}{3}\right) \)
To remove the fractional term, let us choose \( k = 3 \):
\( 3\left(x^2 - \sqrt{2}x + \frac{1}{3}\right) = 3x^2 - 3\sqrt{2}x + 1 \).
In simple words: We write the basic polynomial as \( x^2 - \sqrt{2}x + 1/3 \). To get integer coefficients, we multiply the whole expression by 3.

Exam Tip: Remember that the general form has a negative sign before the sum of zeroes term, and choosing an appropriate constant \( k \) helps clear any denominators.

Level-II
 

Question. If 1 is a zero of the polynomial p(x) = ax2 -3(a-1)x -1, then find the value of a.
Answer: Given that 1 is a zero of the polynomial \( p(x) = ax^2 - 3(a-1)x - 1 \), we substitute \( x = 1 \) and set the expression to zero:
\( p(1) = a(1)^2 - 3(a-1)(1) - 1 = 0 \)
\( a - 3a + 3 - 1 = 0 \)
\( -2a + 2 = 0 \)
\( 2a = 2 \)

\( \implies a = 1 \)
In simple words: Since 1 is a zero, substituting \( x = 1 \) into the polynomial must give an answer of 0. Solving this basic equation gives \( a = 1 \).

Exam Tip: Expand the bracket terms carefully to prevent sign errors when simplifying algebraic equations.

Question. For what value of k, (-4) is zero of the polynomial x2 – x – (2k+2)?
Answer: Let the polynomial be \( p(x) = x^2 - x - (2k + 2) \). Since \( -4 \) is a zero, \( p(-4) = 0 \).
\( (-4)^2 - (-4) - (2k + 2) = 0 \)
\( 16 + 4 - 2k - 2 = 0 \)
\( 18 - 2k = 0 \)
\( 2k = 18 \)

\( \implies k = 9 \)
In simple words: Substitute \( x = -4 \) into the polynomial and set the entire expression to zero. Solving the resulting linear equation for \( k \) gives 9.

Exam Tip: Always place negative numbers in brackets when squaring, i.e., \( (-4)^2 = 16 \), to avoid sign confusion.

Question. Write a quadratic polynomial, the sum and product of whose zeroes are 3 and -2.
Answer: Let the sum of the zeroes be \( S = 3 \) and the product of the zeroes be \( P = -2 \).
The standard equation for a quadratic polynomial is:
\( x^2 - Sx + P \)
Substituting the given values:
\( x^2 - (3)x + (-2) = x^2 - 3x - 2 \).
In simple words: Use the formula \( x^2 - (\text{sum})x + \text{product} \) and plug in 3 and -2 to get \( x^2 - 3x - 2 \).

Exam Tip: The minus sign in front of the middle term is a fixture of the formula. Ensure it is preserved when the sum is positive.

Question. Find the zeroes of the quadratic polynomial 2x2-9-3x and verify the relationship between the zeroes and the coefficients.
Answer: Let the polynomial be rewritten in standard descending form:
\( p(x) = 2x^2 - 3x - 9 \)
To find the zeroes, we set \( p(x) = 0 \) and factor by splitting the middle term:
\( 2x^2 - 6x + 3x - 9 = 0 \)
\( 2x(x - 3) + 3(x - 3) = 0 \)
\( (2x + 3)(x - 3) = 0 \)
Thus, the zeroes are:
\( 2x + 3 = 0 \implies x = -\frac{3}{2} \)
\( x - 3 = 0 \implies x = 3 \)
So, \( \alpha = -\frac{3}{2} \) and \( \beta = 3 \).

Verification of relationship with coefficients:
Here, \( a = 2, b = -3, c = -9 \).
1. Sum of zeroes:
\( \alpha + \beta = -\frac{3}{2} + 3 = \frac{-3 + 6}{2} = \frac{3}{2} \)
From coefficients:
\( -\frac{b}{a} = -\frac{-3}{2} = \frac{3}{2} \)
Thus, \( \alpha + \beta = -\frac{b}{a} \) (Verified).

2. Product of zeroes:
\( \alpha\beta = \left(-\frac{3}{2}\right)(3) = -\frac{9}{2} \)
From coefficients:
\( \frac{c}{a} = -\frac{9}{2} \)
Thus, \( \alpha\beta = \frac{c}{a} \) (Verified).
In simple words: Rearrange the polynomial as \( 2x^2 - 3x - 9 \), then factor it to find the zeroes \( 3 \) and \( -3/2 \). Finally, show that their sum and product match the ratios of the coefficients.

Exam Tip: Always rearrange the polynomial terms into standard format \( ax^2 + bx + c \) before identifying coefficients \( a, b, \) and \( c \).

Question. Write the polynomial whose zeroes are 2 +\(\sqrt{3}\) and 2 - \(\sqrt{3}\).
Answer: Let the zeroes be \( \alpha = 2 + \sqrt{3} \) and \( \beta = 2 - \sqrt{3} \).
We calculate the sum of the zeroes:
\( \alpha + \beta = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4 \)
We calculate the product of the zeroes using the identity \( (a-b)(a+b) = a^2 - b^2 \):
\( \alpha\beta = (2 + \sqrt{3})(2 - \sqrt{3}) = (2)^2 - (\sqrt{3})^2 = 4 - 3 = 1 \)
The quadratic polynomial is:
\( p(x) = x^2 - (\alpha + \beta)x + \alpha\beta \)
Substituting the values:
\( p(x) = x^2 - 4x + 1 \).
In simple words: Add the zeroes to get 4, and multiply them to get 1. Put these values into the standard polynomial formula.

Exam Tip: The product of conjugate irrational roots like \( a + \sqrt{b} \) and \( a - \sqrt{b} \) always results in a rational number.

Level-III
 

Question. Find all the zeroes of the polynomial 2x3+x2-6x-3, if two of its zeroes are -\(\sqrt{3}\) and \(\sqrt{3}\).
Answer: Let the polynomial be \( p(x) = 2x^3 + x^2 - 6x - 3 \).
Since \( -\sqrt{3} \) and \( \sqrt{3} \) are zeroes, the product of their corresponding factors is a factor of \( p(x) \):
\( (x + \sqrt{3})(x - \sqrt{3}) = x^2 - 3 \)
We perform long division of \( 2x^3 + x^2 - 6x - 3 \) by \( x^2 - 3 \) to find the remaining linear factor:
\( \frac{2x^3 + x^2 - 6x - 3}{x^2 - 3} = 2x + 1 \)
To find the third zero, we set this quotient to zero:
\( 2x + 1 = 0 \)

\( \implies x = -\frac{1}{2} \)
Thus, the zeroes of the polynomial are \( -\sqrt{3} \), \( \sqrt{3} \), and \( -\frac{1}{2} \).
In simple words: We multiply the factors of the two known zeroes to get \( x^2 - 3 \). Dividing the main polynomial by this gives us the final factor \( 2x + 1 \), which yields \( -1/2 \).

Exam Tip: Alternatively, you can use the sum of roots formula \( \alpha + \beta + \gamma = -\frac{b}{a} \) to find the third root directly: \( \sqrt{3} + (-\sqrt{3}) + \gamma = -\frac{1}{2} \implies \gamma = -\frac{1}{2} \).

Question. If the polynomial x4+2x3+8x2+12x+18 is divided by another polynomial x2+5, the remainder comes out to be px+q. Find the value of p and q.
Answer: We perform long division of \( x^4 + 2x^3 + 8x^2 + 12x + 18 \) by \( x^2 + 5 \):
First division step:
\( x^2(x^2 + 5) = x^4 + 5x^2 \)
Subtracting this leaves:
\( 2x^3 + 3x^2 + 12x + 18 \)
Second step:
\( 2x(x^2 + 5) = 2x^3 + 10x \)
Subtracting this leaves:
\( 3x^2 + 2x + 18 \)
Third step:
\( 3(x^2 + 5) = 3x^2 + 15 \)
Subtracting this leaves the remainder:
\( 2x + 3 \)
Comparing the remainder \( 2x + 3 \) with the given expression \( px + q \):
\( p = 2 \)
\( q = 3 \).
In simple words: Divide the fourth-degree expression by \( x^2 + 5 \) using long division. The final remaining expression is \( 2x + 3 \), which tells us \( p = 2 \) and \( q = 3 \).

Exam Tip: Be sure to keep terms with similar powers aligned during division to avoid adding unlike algebraic terms.

Question. If the polynomial 6x4+8x3+17x2+21x+7 is divided by another polynomial 3x2+4x+1, the remainder comes out to be (ax+b), find a and b.
Answer: We perform long division of \( 6x^4 + 8x^3 + 17x^2 + 21x + 7 \) by \( 3x^2 + 4x + 1 \):
First division step:
\( 2x^2(3x^2 + 4x + 1) = 6x^4 + 8x^3 + 2x^2 \)
Subtracting this leaves:
\( 15x^2 + 21x + 7 \)
Second step:
\( 5(3x^2 + 4x + 1) = 15x^2 + 20x + 5 \)
Subtracting this leaves the remainder:
\( x + 2 \)
Comparing the remainder \( x + 2 \) with the given expression \( ax + b \):
\( a = 1 \)
\( b = 2 \).
In simple words: Divide the main polynomial by the divisor using long division. Since the leftover remainder is \( x + 2 \), comparing this to \( ax + b \) shows that \( a \) is 1 and \( b \) is 2.

Exam Tip: Be vigilant with signs when performing the subtraction step at the end of each division loop.

Question. If two zeroes of the polynomial f(x)= x3-4x2-3x+12 are \(\sqrt{3}\) and -\(\sqrt{3}\), then find its third zero.
Answer: Let the zeroes of the cubic polynomial be \( \alpha = \sqrt{3} \), \( \beta = -\sqrt{3} \), and the third zero be \( \gamma \).
For the polynomial \( f(x) = x^3 - 4x^2 - 3x + 12 \), the coefficient of \( x^2 \) is \( -4 \) and the coefficient of \( x^3 \) is 1.
Using the sum of zeroes relationship for a cubic polynomial:
\( \alpha + \beta + \gamma = -\frac{\text{Coefficient of } x^2}{\text{Coefficient of } x^3} \)
Substituting our values:
\( \sqrt{3} + (-\sqrt{3}) + \gamma = -\frac{-4}{1} \)
\( 0 + \gamma = 4 \)

\( \implies \gamma = 4 \)
Thus, the third zero is 4.
In simple words: The sum of all three zeroes of this polynomial must equal 4. Since the two given zeroes cancel each other out, the third zero is simply 4.

Exam Tip: Using the sum of zeroes relation is the quickest way to find a single missing zero of a cubic equation, saving valuable exam time.

Question. If \(\alpha\), \(\beta\) are zeroes of the polynomial x2-2x-15 then form a quadratic polynomial whose zeroes are (2\(\alpha\)) and (2\(\beta\)).
Answer: For the polynomial \( x^2 - 2x - 15 \), the sum and product of zeroes are:
\( \alpha + \beta = -\frac{-2}{1} = 2 \)
\( \alpha\beta = \frac{-15}{1} = -15 \)
We need to find a new polynomial with zeroes \( \alpha' = 2\alpha \) and \( \beta' = 2\beta \).
The sum of the new zeroes is:
\( S' = 2\alpha + 2\beta = 2(\alpha + \beta) = 2(2) = 4 \)
The product of the new zeroes is:
\( P' = (2\alpha)(2\beta) = 4\alpha\beta = 4(-15) = -60 \)
The new polynomial is given by:
\( x^2 - S'x + P' = x^2 - 4x - 60 \).
In simple words: Find the original sum and product of zeroes, double them appropriately for the new roots, and plug the results back into the quadratic formula.

Exam Tip: An alternative method is substitution: replace \( x \) with \( \frac{x}{2} \) in the original equation to obtain the new polynomial directly.

Level-IV
 

Question. Find other zeroes of the polynomial p(x)=2x4 +7x3-19x2-14x +30 if two of its zeroes are \(\sqrt{2}\) and -\(\sqrt{2}\).
Answer: Since \( \sqrt{2} \) and \( -\sqrt{2} \) are zeroes, their corresponding factor is:
\( (x - \sqrt{2})(x + \sqrt{2}) = x^2 - 2 \)
We divide \( 2x^4 + 7x^3 - 19x^2 - 14x + 30 \) by \( x^2 - 2 \):
First division step:
\( 2x^2(x^2 - 2) = 2x^4 - 4x^2 \)
Subtracting this leaves:
\( 7x^3 - 15x^2 - 14x + 30 \)
Second step:
\( 7x(x^2 - 2) = 7x^3 - 14x \)
Subtracting this leaves:
\( -15x^2 + 30 \)
Third step:
\( -15(x^2 - 2) = -15x^2 + 30 \)
Subtracting this leaves a remainder of 0.
The quadratic quotient is \( 2x^2 + 7x - 15 \). Setting this to zero to find the other zeroes:
\( 2x^2 + 10x - 3x - 15 = 0 \)
\( 2x(x + 5) - 3(x + 5) = 0 \)
\( (2x - 3)(x + 5) = 0 \)
This gives:
\( x = \frac{3}{2} \) and \( x = -5 \).
Thus, the other zeroes are \( \frac{3}{2} \) and \( -5 \).
In simple words: Divide the starting polynomial by \( x^2 - 2 \) to get the remaining quadratic expression \( 2x^2 + 7x - 15 \). Factoring this gives the final zeroes \( 3/2 \) and \( -5 \).

Exam Tip: Be sure to write the polynomial in proper descending order, using zero placeholders for any missing degrees of \( x \) to keep calculations neat.

Question. Divide 30x4 +11x3-82x2-12x-48 by (3x2 +2x-4) and verify the result by division algorithm.
Answer: We divide the dividend \( f(x) = 30x^4 + 11x^3 - 82x^2 - 12x - 48 \) by the divisor \( g(x) = 3x^2 + 2x - 4 \):
First division step:
\( 10x^2(3x^2 + 2x - 4) = 30x^4 + 20x^3 - 40x^2 \)
Subtracting this leaves:
\( -9x^3 - 42x^2 - 12x - 48 \)
Second step:
\( -3x(3x^2 + 2x - 4) = -9x^3 - 6x^2 + 12x \)
Subtracting this leaves:
\( -36x^2 - 24x - 48 \)
Third step:
\( -12(3x^2 + 2x - 4) = -36x^2 - 24x + 48 \)
Subtracting this leaves the remainder:
\( -96 \)
Thus:
Quotient \( q(x) = 10x^2 - 3x - 12 \)
Remainder \( r(x) = -96 \)

Verification by Division Algorithm:
\( \text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder} \)
\( \text{LHS} = (3x^2 + 2x - 4)(10x^2 - 3x - 12) + (-96) \)
\( = 3x^2(10x^2 - 3x - 12) + 2x(10x^2 - 3x - 12) - 4(10x^2 - 3x - 12) - 96 \)
\( = (30x^4 - 9x^3 - 36x^2) + (20x^3 - 6x^2 - 24x) - (40x^2 - 12x - 48) - 96 \)
\( = 30x^4 + (-9x^3 + 20x^3) + (-36x^2 - 6x^2 - 40x^2) + (-24x + 12x) + (48 - 96) \)
\( = 30x^4 + 11x^3 - 82x^2 - 12x - 48 \)
Since the LHS equals the RHS, the division algorithm is verified.
In simple words: Divide the polynomial step-by-step to find the quotient and remainder. Multiply the divisor by the quotient and add the remainder to show that it matches the original polynomial.

Exam Tip: When performing algebraic expansion during verification, distribute terms carefully and group like-terms systematically.

Question. If the polynomial 6x4 +8x3-5x2+ax+b is exactly divisible by the polynomial 2x2-5, then find the value of a and b.
Answer: We divide the polynomial \( f(x) = 6x^4 + 8x^3 - 5x^2 + ax + b \) by \( g(x) = 2x^2 - 5 \):
First division step:
\( 3x^2(2x^2 - 5) = 6x^4 - 15x^2 \)
Subtracting this leaves:
\( 8x^3 + 10x^2 + ax + b \)
Second step:
\( 4x(2x^2 - 5) = 8x^3 - 20x \)
Subtracting this leaves:
\( 10x^2 + (a + 20)x + b \)
Third step:
\( 5(2x^2 - 5) = 10x^2 - 25 \)
Subtracting this leaves the remainder:
\( (a + 20)x + (b + 25) \)

Since the polynomial is exactly divisible, the remainder must be 0:
\( (a + 20)x + (b + 25) = 0 \)
This means:
\( a + 20 = 0 \implies a = -20 \)
\( b + 25 = 0 \implies b = -25 \).
In simple words: Divide the polynomial by \( 2x^2 - 5 \). Since there should be no remainder, set both parts of the remaining expression to 0 to find \( a = -20 \) and \( b = -25 \).

Exam Tip: Any linear remainder term \( px + q \) must equal 0, meaning both coefficients \( p \) and \( q \) must be set to 0 individually.

Question. Obtain all other zeroes of 3x4 -15x3+13x2+25x-30, if two of its zeroes are \(\sqrt{5/3}\) and -\(\sqrt{5/3}\).
Answer: Since the two zeroes are \( \sqrt{\frac{5}{3}} \) and \( -\sqrt{\frac{5}{3}} \), their product factor is:
\( \left(x - \sqrt{\frac{5}{3}}\right)\left(x + \sqrt{\frac{5}{3}}\right) = x^2 - \frac{5}{3} \)
To avoid working with fractions, we can use the factor \( 3x^2 - 5 \).
Dividing \( 3x^4 - 15x^3 + 13x^2 + 25x - 30 \) by \( 3x^2 - 5 \):
First division step:
\( x^2(3x^2 - 5) = 3x^4 - 5x^2 \)
Subtracting this leaves:
\( -15x^3 + 18x^2 + 25x - 30 \)
Second step:
\( -5x(3x^2 - 5) = -15x^3 + 25x \)
Subtracting this leaves:
\( 18x^2 - 30 \)
Third step:
\( 6(3x^2 - 5) = 18x^2 - 30 \)
Subtracting this leaves a remainder of 0.

The quotient is \( x^2 - 5x + 6 \). Setting this to 0 to find the remaining zeroes:
\( x^2 - 5x + 6 = 0 \)
\( x^2 - 2x - 3x + 6 = 0 \)
\( x(x - 2) - 3(x - 2) = 0 \)
\( (x - 2)(x - 3) = 0 \)
This gives:
\( x = 2 \) and \( x = 3 \).
Thus, the other zeroes are 2 and 3.
In simple words: The given zeroes combine to give the factor \( 3x^2 - 5 \). Dividing by this factor leaves us with \( x^2 - 5x + 6 \), which splits to give the remaining zeroes 2 and 3.

Exam Tip: Multiplying the fractional factor \( x^2 - \frac{a}{b} \) by the denominator \( b \) simplifies long division by getting rid of fractions.

Question. If \(\alpha\), \(\beta\) are zeroes of the quadratic polynomial p(x)=kx2+4x+4 such that \(\alpha\)2 +\(\beta\)2=24, find the value of k.
Answer: For the quadratic polynomial \( p(x) = kx^2 + 4x + 4 \), the coefficients are \( a = k, b = 4, c = 4 \).
The sum of the zeroes is:
\( \alpha + \beta = -\frac{b}{a} = -\frac{4}{k} \)
The product of the zeroes is:
\( \alpha\beta = \frac{c}{a} = \frac{4}{k} \)
We are given:
\( \alpha^2 + \beta^2 = 24 \)
Using the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \):
\( \left(-\frac{4}{k}\right)^2 - 2\left(\frac{4}{k}\right) = 24 \)
\( \frac{16}{k^2} - \frac{8}{k} = 24 \)
Dividing by 8:
\( \frac{2}{k^2} - \frac{1}{k} = 3 \)
Multiplying through by \( k^2 \):
\( 2 - k = 3k^2 \)
\( 3k^2 + k - 2 = 0 \)
Factoring the quadratic equation:
\( 3k^2 + 3k - 2k - 2 = 0 \)
\( 3k(k + 1) - 2(k + 1) = 0 \)
\( (3k - 2)(k + 1) = 0 \)
This gives:
\( 3k - 2 = 0 \implies k = \frac{2}{3} \)
\( k + 1 = 0 \implies k = -1 \).
Thus, the values of \( k \) are \( \frac{2}{3} \) and \( -1 \).
In simple words: Write \( \alpha^2 + \beta^2 \) using the sum and product formulas. This results in a quadratic equation in terms of \( k \) which we solve to get \( k = 2/3 \) and \( k = -1 \).

Exam Tip: Be sure to multiply all terms by \( k^2 \) to simplify rational equations, and double-check your factorization of the resulting quadratic.

Self Evaluation
 

Question. If \(\alpha\), \(\beta\) are zeroes of the quadratic polynomial ax2+bx+c then find (a) \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\) (b) \(\alpha^2 + \beta^2\)
Answer: For the quadratic polynomial \( ax^2 + bx + c \), we have:
\( \alpha + \beta = -\frac{b}{a} \)
\( \alpha\beta = \frac{c}{a} \)

(a) To find \( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \):
\( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} \)
Using the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \):
\( \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta} = \frac{\left(-\frac{b}{a}\right)^2 - 2\left(\frac{c}{a}\right)}{\frac{c}{a}} \)
\( = \frac{\frac{b^2}{a^2} - \frac{2c}{a}}{\frac{c}{a}} = \frac{\frac{b^2 - 2ac}{a^2}}{\frac{c}{a}} \)
\( = \frac{b^2 - 2ac}{a^2} \times \frac{a}{c} = \frac{b^2 - 2ac}{ac} \)

(b) To find \( \alpha^2 + \beta^2 \):
\( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \)
Substituting the values:
\( \alpha^2 + \beta^2 = \left(-\frac{b}{a}\right)^2 - 2\left(\frac{c}{a}\right) = \frac{b^2}{a^2} - \frac{2c}{a} = \frac{b^2 - 2ac}{a^2} \).
In simple words: Express the required fractional and squared terms using the sum \( -b/a \) and product \( c/a \) equations, then simplify the algebraic fractions.

Exam Tip: Practice symmetric expressions in \( \alpha \) and \( \beta \) regularly as they are essential for scoring well in polynomial algebra.

Question. If \(\alpha\), \(\beta\) are zeroes of the quadratic polynomial ax2+bx+c then find the value of \(\alpha^2 - \beta^2\)
Answer: For the polynomial \( ax^2 + bx + c \), we have:
\( \alpha + \beta = -\frac{b}{a} \)
\( \alpha\beta = \frac{c}{a} \)
Using the difference of squares identity:
\( \alpha^2 - \beta^2 = (\alpha + \beta)(\alpha - \beta) \)
To find \( \alpha - \beta \), we use:
\( (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \)
\( (\alpha - \beta)^2 = \left(-\frac{b}{a}\right)^2 - 4\left(\frac{c}{a}\right) = \frac{b^2}{a^2} - \frac{4c}{a} = \frac{b^2 - 4ac}{a^2} \)
Taking the square root:
\( \alpha - \beta = \pm \frac{\sqrt{b^2 - 4ac}}{a} \)
Substituting this back into the formula for \( \alpha^2 - \beta^2 \):
\( \alpha^2 - \beta^2 = \left(-\frac{b}{a}\right)\left(\pm \frac{\sqrt{b^2 - 4ac}}{a}\right) = \mp \frac{b\sqrt{b^2 - 4ac}}{a^2} \).
In simple words: Rewrite the expression as \( (\alpha + \beta)(\alpha - \beta) \). Find the value of \( \alpha - \beta \) using the square root of \( (\alpha+\beta)^2 - 4\alpha\beta \), and multiply them together.

Exam Tip: Remember to write the \( \pm \) or \( \mp \) symbol since the value of the difference can be positive or negative depending on which zero is larger.

Question. If \(\alpha\), \(\beta\) are zeroes of the quadratic polynomial ax2+bx+c then find the value of \(\alpha^3 + \beta^3\)
Answer: For the quadratic polynomial \( ax^2 + bx + c \), the sum and product of zeroes are:
\( \alpha + \beta = -\frac{b}{a} \)
\( \alpha\beta = \frac{c}{a} \)
Using the algebraic identity for sum of cubes:
\( \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \)
We rewrite the second bracket in terms of sum and product:
\( \alpha^3 + \beta^3 = (\alpha + \beta)[(\alpha + \beta)^2 - 3\alpha\beta] \)
Substituting our values:
\( \alpha^3 + \beta^3 = \left(-\frac{b}{a}\right) \left[ \left(-\frac{b}{a}\right)^2 - 3\left(\frac{c}{a}\right) \right] \)
\( = \left(-\frac{b}{a}\right) \left[ \frac{b^2}{a^2} - \frac{3c}{a} \right] \)
\( = \left(-\frac{b}{a}\right) \left[ \frac{b^2 - 3ac}{a^2} \right] \)
\( = \frac{-b(b^2 - 3ac)}{a^3} = \frac{3abc - b^3}{a^3} \).
In simple words: Express \( \alpha^3 + \beta^3 \) as \( (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta) \) and substitute the values of the sum and product of roots.

Exam Tip: Expanding \( \alpha^3 + \beta^3 \) directly as \( (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) \) is often faster and less prone to mistakes.

Question. What must be added to 6x5+5x4+11x3-3x2+x+5 so that it may be exactly divisible by 3x2-2x+4?
Answer: We divide the polynomial \( f(x) = 6x^5 + 5x^4 + 11x^3 - 3x^2 + x + 5 \) by \( g(x) = 3x^2 - 2x + 4 \) using long division:
First division step:
\( 2x^3(3x^2 - 2x + 4) = 6x^5 - 4x^4 + 8x^3 \)
Subtracting this leaves:
\( 9x^4 + 3x^3 - 3x^2 + x + 5 \)
Second step:
\( 3x^2(3x^2 - 2x + 4) = 9x^4 - 6x^3 + 12x^2 \)
Subtracting this leaves:
\( 9x^3 - 15x^2 + x + 5 \)
Third step:
\( 3x(3x^2 - 2x + 4) = 9x^3 - 6x^2 + 12x \)
Subtracting this leaves:
\( -9x^2 - 11x + 5 \)
Fourth step:
\( -3(3x^2 - 2x + 4) = -9x^2 + 6x - 12 \)
Subtracting this leaves the final remainder:
\( -17x + 17 \)

To make the polynomial exactly divisible, we must add the negative of the remainder:
\( -(-17x + 17) = 17x - 17 \).
Therefore, \( 17x - 17 \) must be added.
In simple words: Divide the main polynomial by \( 3x^2 - 2x + 4 \). Find the leftover remainder, which is \( -17x + 17 \), and change its signs to find what must be added.

Exam Tip: Remember: the term to be subtracted is the remainder \( r(x) \), while the term to be added is always the negative remainder \( -r(x) \).

Question. If the square of difference of the zeroes of the quadratic polynomial f(x)=x2+px+45 is equal to 144, find the value of p.
Answer: Let the zeroes of \( f(x) = x^2 + px + 45 \) be \( \alpha \) and \( \beta \).
From the coefficients, we have:
\( \alpha + \beta = -p \)
\( \alpha\beta = 45 \)
We are given:
\( (\alpha - \beta)^2 = 144 \)
Using the identity:
\( (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \)
Substituting our values:
\( 144 = (-p)^2 - 4(45) \)
\( 144 = p^2 - 180 \)
\( p^2 = 144 + 180 \)
\( p^2 = 324 \)
Taking the square root on both sides:
\( p = \pm 18 \).
In simple words: Use the relationship \( (\alpha-\beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta \) to set up an equation for \( p \). Solving this equation gives \( p = 18 \) or \( p = -18 \).

Exam Tip: Never omit the negative square root when taking roots of a variable squared; write both \( \pm \) values to earn full marks.