CBSE Class 10 Maths HOTs Polynomials Set D

Very Short Answer Type Questions

Question. Form a quadratic polynomial, the sum and product of whose zeroes are (–3) and 2 respectively.
Answer: A general form of a quadratic polynomial is \( ax^2 + bx + c \)
Here, \( \alpha + \beta = -\frac{b}{a} = -3 \) and \( \alpha\beta = \frac{c}{a} = 2 \)
where, \( \alpha \) and \( \beta \) are the roots of the given polynomial.
So, the required polynomial is \( x^2 + 3x + 2 \).

Question. Find the value of k for which the roots of the equation \( 3x^2 - 10x + k = 0 \) are reciprocal of each other.
Answer: Given, equation is \( 3x^2 – 10x + k = 0 \), where roots are reciprocals of each other.
Let the roots be \( \alpha \) and \( \frac{1}{\alpha} \)
\( \therefore \) Product of roots = \( \frac{c}{a} \)
\( \Rightarrow \alpha \cdot \frac{1}{\alpha} = \frac{k}{3} \) [\( \because a = 3, b = -10, c = k \)]
\( \Rightarrow 1 = \frac{k}{3} \)
\( \Rightarrow k = 3 \)

Question. Determine the degree of the polynomial \( (x + 1) (x^2 – x – x^4 + 1) \).
Answer: Given polynomial in standard form is:
\( -x^5 – x^4 + x^3 + 1 \)
So, its degree is 5.

Question. If the product of two zeros of the polynomial \( p(x) = 2x^3 + 6x^2 – 4x + 9 \) is 3, find the third zero of the polynomial.
Answer: If \( \alpha, \beta, \) and \( \gamma \) be the three zeros of \( p(x) \). Then,
\( \alpha\beta\gamma = -\frac{9}{2} \)
Since, \( \alpha\beta = 3 \),
we get \( \gamma = -\frac{9}{2} \times \frac{1}{3} = -\frac{3}{2} \)
Thus, the third zero of \( p(x) \) is \( -\frac{3}{2} \).

Question. If \( \alpha \) and \( \beta \) are the zeros of the polynomial \( p(x) = 4x^2 – 2x - 3 \), find the value of \( \frac{1}{\alpha} + \frac{1}{\beta} \).
Answer: Here, \( \alpha + \beta = \frac{2}{4} \) or \( \frac{1}{2} \) and \( \alpha\beta = \frac{-3}{4} \).
So, \( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{1/2}{-3/4} = -\frac{2}{3} \).

Question. If one of the zeros of polynomial \( p(x) = (k - 1)x^2 – kx + 1 \) is -3, find the value of k.
Answer: Since, (–3) is a zero of \( p(x) \), we have,
\( (k – 1)(-3)^2 – k(-3) + 1 = 0 \)
\( \Rightarrow 9k – 9 + 3k + 1 = 0 \)
\( \Rightarrow 12k = 8 \Rightarrow k = \frac{2}{3} \)

Question. If \( \alpha \) and \( \beta \) be the roots of the equation \( x^2 - 1 = 0 \), then show that \( \frac{1}{\alpha} + \frac{1}{\beta} = 0 \).
Answer: Here, \( \alpha + \beta = \frac{0}{1} = 0 \)
\( [\because \text{Sum of roots} = \frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} ] \)
Also, \( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{0}{-1} = 0 \)
Thus, \( \frac{1}{\alpha} + \frac{1}{\beta} = 0 \).

Question. A teacher asked 10 of his students to write a polynomial in one variable on a paper and then to handover the paper. The following were the answer given by the students:
\( 2x + 3, 3x^2 + 7x + 2, 4x^3 + 3x^2 + 2, x^3 + \sqrt{3x} + 7, 7x + \sqrt{7}, 5x^3 – 7x + 2, 2x^3 + 3 – \frac{5}{x}, 5x – \frac{1}{2}, ax^3 + bx^2 + cx + d, x + \frac{1}{x} \).
Answer the following questions:
(A) How many of the above ten are not polynomials?
(B) How many of the above ten are quadratic polynomials?
Answer: (A) Three, namely: \( x^3 + \sqrt{3x} + 7 \), \( 2x^3 + 3 - \frac{5}{x} \), \( x + \frac{1}{x} \)
(As they contain square roots of the variable and negative power of x)
(B) One, namely \( 3x^2 + 7x + 2 \)

Question. If one of the zeroes of the quadratic polynomial \( f(x) = 4x^2 – 8kx – 9 \) is equal in magnitude but opposite in sign of the other, then find the value of k.
Answer: \( f(x) = 4x^2 – 8kx – 9 \)
Let one of the zeroes of the polynomial be \( \alpha \) and the other zeroes be \( -\alpha \)
Sum of zeroes = \( -\left(\frac{b}{a}\right) = \frac{8k}{4} \)
\( \alpha + (-\alpha) = 0 \)
So, \( \frac{8k}{4} = 0 \Rightarrow k = 0 \)

Question. Can \( (x – 5) \) be the remainder on division of a polynomial \( p(x) \) by \( (x + 8) \)?
Answer: No. We know that we cannot divide the polynomials which have same degree.
As we can see that degree of \( (x – 5) \) = degree of \( (x + 8) \)
So, they are not divisible.

Short Answer Type Questions

Question. If the zeros of the polynomial \( x^3 – 3x^2 + x + 1 \) are \( a – b, a \) and \( a + b \), then find the values of a and b.
Answer: As \( (a – b), a \) and \( (a + b) \) are zeros of \( x^3 – 3x^2 + x + 1 \), we have:
\( (a – b) + a + (a + b) = 3 \)
\( \Rightarrow 3a = 3 \), or \( a = 1 \) ...(i)
\( a (a – b) + a (a + b) + (a – b) (a + b) = 1 \)
\( \Rightarrow 3a^2 – b^2 = 1 \) ...(ii)
and \( (a – b) a (a + b) = -1 \)
\( \Rightarrow a(a^2 – b^2) = -1 \) ...(iii)
From (i) and (ii), we have \( b = \pm \sqrt{2} \)
Thus, \( a = 1, b = \pm \sqrt{2} \)

Question. What number should be added to the polynomial \( x^2 – 5x + 4 \) so that 3 is the zero of the polynomial?
Answer: Let k be the number to be added to the given polynomial. Then the polynomial becomes
\( x^2 – 5x + (4 + k) \)
As 3 is the zero of the polynomial, we get:
\( (3)^2 – 5(3) + (4 + k) = 0 \)
\( \Rightarrow (4 + k) = 15 – 9 \)
\( \Rightarrow 4 + k = 6 \)
\( \Rightarrow k = 2 \)
Thus, 2 is to be added to the polynomial.

Question. If the zeroes of a polynomial \( x^2 – 8x + k = 0 \), is the HCF of (6, 12), then find the value of k.
Answer: HCF of (6, 12) = 6
So, 6 is one of the roots of the polynomial.
\( f(x) = x^2 – 8x + k = 0 \)
\( f(6) = (6)^2 – 8(6) + k = 0 \)
\( 36 – 48 + k = 0 \)
\( -12 + k = 0 \Rightarrow k = 12 \).

Question. Find the quadratic polynomial sum and product of whose zeroes are – 1 and – 20 respectively. Also, find the zeroes of the polynomial so obtained.
Answer: Let \( \alpha \) and \( \beta \) be the zeroes of the polynomial.
Given : sum of zeroes, \( \alpha + \beta = – 1 \)
product of zeroes, \( \alpha\beta = – 20 \)
Equation of polynomial :
\( x^2 – (\text{sum of zeroes}) x + \text{product of zeroes} = 0 \)
\( \therefore x^2 – (– 1)x + (– 20) = 0 \)
\( \Rightarrow x^2 + x – 20 = 0 \)
On splitting the middle term,
\( x^2 + 5x – 4x – 20 = 0 \)
\( \Rightarrow x(x + 5) – 4(x + 5) = 0 \)
\( \Rightarrow (x – 4) (x + 5) = 0 \)
\( \Rightarrow x = 4, – 5 \)
Hence, the zeroes of the polynomial are 4 and –5.

Question. Find a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial \( f(x) = ax^2+ bx + c, a \neq 0, c \neq 0 \).
Answer: Let \( \alpha, \beta \) be the zeroes of \( f(x) = ax^2 + bx + c \). Thus
\( \alpha + \beta = -\frac{b}{a} \) and \( \alpha\beta = \frac{c}{a} \)
Now, \( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-b/a}{c/a} = -\frac{b}{c} \)
\( \frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha\beta} = \frac{1}{c/a} = \frac{a}{c} \)
\( \therefore \) Polynomial is: \( x^2 – (\text{sum of roots}) x + \text{product roots} \)
\( x^2 - \left(-\frac{b}{c}\right)x + \frac{a}{c} = 0 \)
\( \Rightarrow cx^2 + bx + a = 0 \)
So, the required polynomial is \( cx^2 + bx + a \).

Question. If the zeroes of the polynomial \( x^2 + px + q \) are double the value to the zeroes of \( 2x^2 – 5x – 3 \), find the value of p and q.
Answer: Let \( \alpha \) and \( \beta \) are zeroes of the \( 2x^2 – 5x – 3 \)
\( \alpha + \beta = -\frac{b}{a} = \frac{5}{2} \) ...(i)
\( \alpha\beta = \frac{c}{a} = -\frac{3}{2} \) ...(ii)
According to the question,
\( 2\alpha \) and \( 2\beta \) are zeroes of \( x^2 + px + q \)
\( 2\alpha + 2\beta = -p \Rightarrow 2(\alpha + \beta) = -p \)
\( 2 \times \left(\frac{5}{2}\right) = -p \) [from eqn. (i)]
\( p = -5 \)
\( 2\alpha \times 2\beta = q \Rightarrow 4\alpha\beta = q \)
\( 4 \times \left(-\frac{3}{2}\right) = q \) [from eqn. (ii)]
\( q = -6 \)
Hence, \( p = -5 \) and \( q = -6 \).

Question. Find the value of k such that the polynomial \( x^2 – (k + 6)x + 2(2k – 1) \) has the sum of its zeros equal to half of their product.
Answer: Given polynomial is :
\( p(x) = x^2 – (k + 6) x + 2(2k – 1) \)
In the given quadratic equation:
\( a = 1 \)
\( b = – (k + 6) \)
\( c = 2(2k – 1) \)
Sum of zeroes = \( -\frac{b}{a} = k + 6 \) ...(i)
Product of zeroes = \( \frac{c}{a} = 2(2k – 1) \) ...(ii)
According to the given condition:
Sum of the zeroes = \( \frac{1}{2} \times \) Product of zeroes
\( \Rightarrow k + 6 = \frac{1}{2} \times 2 (2k – 1) \)
\( \Rightarrow k + 6 = 2k – 1 \)
\( \Rightarrow 2k – k = 6 + 1 \)
\( \Rightarrow k = 7 \)
Hence, the value of k is 7.

Question. Find a quadratic polynomial whose zeroes are 1 and -3. Verify the relation between the coefficients and zeroes of polynomial.
Answer: Sum of zeroes,
\( S = 1 + (–3) = –2 \) ...(i)
Product of zeroes, \( P = 1 \times (–3) = –3 \) ...(ii)
Quadratic polynomial
\( p(x) = x^2 – Sx + P \)
\( = x^2 – (–2)x – 3 = x^2 + 2x – 3 \)
Here, \( a = 1, b = 2, c = –3 \)
\( -\frac{b}{a} = -\frac{2}{1} = -2 \)
Sum of zeroes = \( -\frac{b}{a} = -2 \neq 2 \) (Wait, sum from (i) is -2. \( -b/a = -2/1 = -2 \). Matches.) [Corrected: Sum = -2, Formula gives -2]
Also, \( \frac{c}{a} = -\frac{3}{1} = -3 \)
Product of zeroes = \( \frac{c}{a} = -3 \) [using eqn. (ii)]
Hence, verified.

Question. If one root of the equation \( 3x^2 - 8x + 2k + 1 = 0 \) is seven times the other, find the two roots and the value of k.
Answer: Let \( \alpha \) and \( 7\alpha \) be the two roots of the equation:
\( 3x^2 – 8x + (2k + 1) = 0 \)
Then, \( \alpha + 7\alpha = 8\alpha = \frac{8}{3} \) .....(i)
and \( \alpha \cdot (7\alpha) = 7\alpha^2 = \frac{2k + 1}{3} \) .....(ii)
From (i) \( \alpha = \frac{1}{3} \). So, the two roots are \( \frac{1}{3} \) and \( \frac{7}{3} \)
Using \( \alpha = \frac{1}{3} \) in (ii), we have:
\( 7\left(\frac{1}{3}\right)^2 = \frac{2k + 1}{3} \)
\( \Rightarrow 2k + 1 = \frac{7}{3} \)
\( \Rightarrow 2k = \frac{4}{3} \)
\( \Rightarrow k = \frac{2}{3} \)

Question. Without actually calculating the zeroes, form a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial \( 5x^2 + 2x - 3 \).
Answer: Let \( \alpha \) and \( \beta \) be the zeroes of \( 5x^2 + 2x – 3 \)
Then, \( \alpha + \beta = -\frac{b}{a} = -\frac{2}{5} \) and \( \alpha\beta = \frac{c}{a} = -\frac{3}{5} \)
Now
\( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-2/5}{-3/5} = \frac{2}{3} \)
and
\( \frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha\beta} = \frac{1}{-3/5} = -\frac{5}{3} \)
Thus, a quadratic polynomial where zeroes are \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \) is
\( x^2 – (\text{sum of roots}) x + \text{product of roots} = 0 \)
\( \Rightarrow x^2 - \frac{2}{3}x - \frac{5}{3} = 0 \)
i.e., \( 3x^2 – 2x – 5 \)

Long Short Answer Type Questions

Question. Find the zeroes of following polynomials by factorisation method and verify relation between the zeroes and coefficients of polynomials.
(A) \( 2x^2 + \frac{7}{2}x + \frac{3}{4} \)
(B) \( 2s^2 - (1 + 2\sqrt{2})s + \sqrt{2} \)
(C) \( 7y^2 – \frac{11}{3}y – \frac{2}{3} \)
Answer: (A) Let \( f(x) = 2x^2 + \frac{7}{2}x + \frac{3}{4} \)
\( = 8x^2 + 14x + 3 \) (Multiplying the given equation by 4)
\( = 8x^2 + (12x + 2x) + 3 \)
\( = 8x^2 + 12x + 2x + 3 \)
\( = 4x(2x + 3) + 1(2x + 3) \)
\( = (2x + 3)(4x + 1) \)
The zeroes of f(x) are given by f(x) = 0.
So, the value of \( 2x^2 + \frac{7}{2}x + \frac{3}{4} \) is zero when
\( x = -\frac{3}{2} \) or \( x = -\frac{1}{4} \)
\( \Rightarrow x = -\frac{3}{2}, -\frac{1}{4} \)
Verification:
Sum of the zeroes = – (coefficient of x) ÷ coefficient of \( x^2 \)
\( \alpha + \beta = -\frac{b}{a} \)
\( \left(-\frac{3}{2}\right) + \left(-\frac{1}{4}\right) = -\frac{7}{4} \)
\( -\frac{7}{4} = -\frac{7}{4} \)
Product of the zeroes = constant term ÷ coefficient of \( x^2 \)
\( \alpha\beta = \frac{c}{a} \)
\( \left(-\frac{3}{2}\right) \left(-\frac{1}{4}\right) = \frac{3}{4} \div 2 \)
\( \frac{3}{8} = \frac{3}{8} \)
Hence, verified.

(B) Let \( f(s) = 2s^2 – (1 + 2\sqrt{2})s + \sqrt{2} \)
\( = 2s^2 – s – 2\sqrt{2}s + \sqrt{2} \)
\( = s(2s – 1) – \sqrt{2}(2s – 1) \)
\( = (2s – 1)(s – \sqrt{2}) \)
The zeroes of f(s) are given by f(s) = 0
So, the value is zero when \( 2s^2 – (1 + 2\sqrt{2})s + \sqrt{2} = 0 \)
i.e., when \( s = \frac{1}{2} \) or \( \sqrt{2} \)
\( \Rightarrow s = \frac{1}{2}, \sqrt{2} \)
Verification:
Sum of the zeroes = – (coefficient of s) ÷ coefficient of \( s^2 \)
\( \alpha + \beta = -\frac{b}{a} \)
\( \frac{1}{2} + \sqrt{2} = \frac{-(-(1 + 2\sqrt{2}))}{2} \)
\( \frac{1 + 2\sqrt{2}}{2} = \frac{1 + 2\sqrt{2}}{2} \)
Product of the zeroes = constant term ÷ coefficient of \( s^2 \)
\( \alpha\beta = \frac{c}{a} \)
\( \frac{1}{2} \times \sqrt{2} = \frac{\sqrt{2}}{2} \)
\( \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} \)
Hence, verified.

(C) Let \( f(y) = 7y^2 – \frac{11}{3}y – \frac{2}{3} \)
\( = 21y^2 – 11y – 2 \)
\( = 21y^2 + (3y – 14y) – 2 \)
\( = 21y^2 + 3y – 14y – 2 \)
\( = 3y(7y + 1) – 2(7y + 1) \)
\( = (7y + 1)(3y – 2) \)
The zeroes of f(y) are given by f(y) = 0
So, the value of \( 7y^2 – \frac{11}{3}y – \frac{2}{3} \) is zero when
\( y = -\frac{1}{7} \) or \( y = \frac{2}{3} \)
\( \Rightarrow y = -\frac{1}{7}, \frac{2}{3} \)
Verification:
Sum of the zeroes = – (coefficient of y) ÷ coefficient of \( y^2 \)
\( \alpha + \beta = -\frac{b}{a} \)
\( -\frac{1}{7} + \frac{2}{3} = \frac{-(-11)}{21} \)
\( \frac{11}{21} = \frac{11}{21} \)
Product of the zeroes = constant term ÷ coefficient of \( y^2 \)
\( \alpha\beta = \frac{c}{a} \)
\( \left(-\frac{1}{7}\right) \left(\frac{2}{3}\right) = \frac{-2}{3} \div 7 \)
\( -\frac{2}{21} = -\frac{2}{21} \)
Hence, verified.

Question. Obtain other zeroes of the polynomial \( f(x) = 2x^4 + 3x^3 - 5x^2 - 9x - 3 \) if two of its zeroes are \( \sqrt{3} \) and \( -\sqrt{3} \).
Answer: Since \( \sqrt{3} \) and \( -\sqrt{3} \) are zeroes of \( f(x) \),
\( (x - \sqrt{3})(x + \sqrt{3}) \), i.e., \( (x^2 – 3) \) is a factor of \( f(x) \)
to obtain other two zeroes, we shall determine the quotient, by dividing \( f(x) \) with \( (x^2 – 3) \)
Dividing \( 2x^4 + 3x^3 - 5x^2 - 9x - 3 \) by \( x^2 - 3 \):
Quotient = \( 2x^2 + 3x + 1 \)
\( = (2x + 1) (x + 1) \)
So, the two zeroes are – 1 and \( -\frac{1}{2} \).

Question. Given that the zeroes of the cubic polynomial \( x^3 - 6x^2 + 3x + 10 \) are of the form a, a + b, a + 2b for some real numbers a and b, find the values of a and b as well as the zeroes of the given polynomial.
Answer: Let \( p(x) = x^3 – 6x^2 + 3x + 10 \)
and a, (a + b) and (a + 2b) are the zeroes of \( p(x) \).
We know:
Sum of the zeroes = –(coefficient of \( x^2 \)) ÷ coefficient of \( x^3 \)
\( \Rightarrow a + (a + b) + (a + 2b) = –(–6) \)
\( \Rightarrow 3a + 3b = 6 \)
\( \Rightarrow a + b = 2 \)
\( \Rightarrow a = 2 – b \) ...(i)
Product of all the zeroes = –(constant term) ÷ coefficient of \( x^3 \)
\( a(a + b)(a + 2b) = –10 \)
\( (2 – b) (2) (2 + b) = –10 \) [Using eqn. (i)]
\( (2 – b) (2 + b) = –5 \)
\( 4 – b^2 = – 5 \)
\( \Rightarrow b^2 = 9 \)
\( \Rightarrow b = \pm 3 \)
When b = 3, \( a = 2 – 3 = –1 \) [Using equation (i)]
\( \Rightarrow a = –1 \) when b = 3.
When b = –3, \( a = 2 – (–3) = 5 \) [Using equation (i)]
\( \Rightarrow a = 5 \) when b = –3.
Case 1: when a = –1 and b = 3
The zeroes of the polynomial are: -1, 2, 5.
Case 2: when a = 5 and b = –3
The zeroes of the polynomial are: 5, 2, -1.
In both the cases, the zeroes of the polynomial are –1, 2, 5.

Question. Given that \( \sqrt{2} \) is a zero of the cubic polynomial \( 6x^3 + \sqrt{2}x^2 - 10x - 4\sqrt{2} \), find its other two zeroes.
Answer: Let \( p(x) = 6x^3 + \sqrt{2}x^2 – 10x – 4\sqrt{2} \)
As \( \sqrt{2} \) is one of the zeroes of \( p(x) \).
\( \Rightarrow g(x) = (x – \sqrt{2}) \) is one of the factors of \( p(x) \).
Dividing \( 6x^3 + \sqrt{2}x^2 – 10x – 4\sqrt{2} \) by \( (x – \sqrt{2}) \):
Quotient = \( 6x^2 + 7\sqrt{2}x + 4 \)
Then,
\( \Rightarrow 6x^3 + \sqrt{2}x^2 – 10x – 4\sqrt{2} = (x – \sqrt{2}) (6x^2 + 7\sqrt{2}x + 4) \)
\( = (x – \sqrt{2}) \{6x^2 + (3\sqrt{2}x + 4\sqrt{2}x) + 4\} \) (by splitting the middle term)
\( = (x – \sqrt{2}) \{6x^2 + 3\sqrt{2}x + 4\sqrt{2}x + 4\} \)
\( = (x – \sqrt{2}) \{3\sqrt{2}x (\sqrt{2}x + 1) + 4(\sqrt{2}x + 1)\} \)
\( = (x – \sqrt{2}) (\sqrt{2}x + 1) (3\sqrt{2}x + 4) \)
\( \Rightarrow x = \sqrt{2}, -\frac{1}{\sqrt{2}} \) or \( -\frac{4}{3\sqrt{2}} \)
Thus, the other two zeroes are \( -\frac{1}{\sqrt{2}} \) (or \( -\frac{\sqrt{2}}{2} \)) and \( -\frac{4}{3\sqrt{2}} \) (or \( -\frac{2\sqrt{2}}{3} \)).

Question. Given that \( x - \sqrt{5} \) is a factor of the cubic polynomial \( x^3 - 3\sqrt{5}x^2 + 13x - 3\sqrt{5} \), find all the zeroes of the polynomial.
Answer: Let \( p(x) = x^3 – 3\sqrt{5}x^2 + 13x – 3\sqrt{5} \)
As \( \sqrt{5} \) is one of the zeroes of \( p(x) \).
\( \Rightarrow (x – \sqrt{5}) \) is one of the factors of \( p(x) \).
Dividing \( p(x) \) by \( (x – \sqrt{5}) \):
Quotient = \( x^2 – 2\sqrt{5}x + 3 \)
Now \( p(x) = (x – \sqrt{5})(x^2 – 2\sqrt{5}x + 3) \)
\( = (x – \sqrt{5})[x^2 – \{(\sqrt{5} + \sqrt{2})x + (\sqrt{5} – \sqrt{2})x\} + 3] \)
\( = (x – \sqrt{5})\{x – (\sqrt{5} + \sqrt{2})\} \{x – (\sqrt{5} – \sqrt{2})\} \)
So, all the zeroes of the given polynomial are \( (\sqrt{5} + \sqrt{2}), (\sqrt{5} – \sqrt{2}) \) and \( \sqrt{5} \).

Question. For which values of a and b are the zeroes of \( q(x) = x^3 + 2x^2 + a \) also the zeroes of the polynomial \( p(x) = x^5 - x^4 - 4x^3 + 3x^2 + 3x + b \)?
Answer: Let \( p(x) = x^5 – x^4 – 4x^3 + 3x^2 + 3x + b \) and \( q(x) = x^3 + 2x^2 + a \).
Since, the zeroes of the polynomial \( q(x) \) are also zeroes of \( p(x) \), we can say that \( q(x) \) is a factor of \( p(x) \).
Then, on dividing \( p(x) \) by \( q(x) \):
Quotient = \( x^2 - 3x + 2 \)
Remainder \( r(x) = –(a + 1)x^2 + 3(1 + a)x + b – 2a \)
Since \( q(x) \) is factor of \( p(x) \), Remainder = 0
\( \Rightarrow –(a + 1)x^2 + 3(1 + a)x + b – 2a = 0.x^2 + 0.x + 0 \)
On comparing the coefficients of \( x^2 \) and constant term, we get
\( –(a + 1) = 0 \Rightarrow a = – 1 \)
and \( b – 2a = 0 \Rightarrow b = 2a = 2(–1) = – 2 \)