CBSE Class 10 Maths HOTs Polynomials Set C

Very Short Answer Type Questions

Question. If sum of the zeroes of the quadratic polynomial \(3x^2 - kx + 6\) is 3, then find the value of \(k\).
Answer: We have \(p(x) = 3x^2 - kx - 6\)
Sum of the zeroes \(= 3 = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}\)
Thus \(3 = -\frac{(-k)}{3} \Rightarrow k = 9\)

Question. If -1 is a zero of the polynomial \(f(x) = x^2 - 7x - 8\), then calculate the other zero.
Answer: We have \(f(x) = x^2 - 7x - 8\)
Let other zero be \(k\), then we have
Sum of zeroes, \(-1 + k = -\left(\frac{-7}{1}\right) = 7\)
or \(k = 8\)

Question. On dividing \( x^3 - 5x^2 + 6x + 4 \) by a polynomial \( g(x) \), the quotient and the remainder were \( x - 3 \) and \( 4 \) respectively. Find \( g(x) \).
Answer: We have \( x^3 - 5x^2 + 6x + 4 = g(x)(x - 3) + 4 \Rightarrow g(x) = \frac{x^3 - 5x^2 + 6x + 4 - 4}{x - 3} = \frac{x^3 - 5x^2 + 6x}{x - 3} \). Dividing \( x^3 - 5x^2 + 6x \) by \( x - 3 \), we get \( g(x) = x^2 - 2x \).

Question. Find the quotient and remainder on dividing \( p(x) \) by \( g(x) \): \( p(x) = 4x^3 + 8x^2 + 8x + 7 \); \( g(x) = 2x^2 - x + 1 \).
Answer: Using long division, \( (4x^3 + 8x^2 + 8x + 7) \div (2x^2 - x + 1) \). The quotient is \( 2x + 5 \) and the remainder is \( 11x + 2 \).

Question. Check whether the polynomial \( g(x) = x^2 + 3x + 1 \) is a factor of the polynomial \( f(x) = 3x^4 + 5x^3 - 7x^2 + 2x + 4 \).
Answer: Performing long division of \( f(x) \) by \( g(x) \), we find that the remainder is \( 2 \). Since remainder is not zero, polynomial \( g(x) \) is not a factor of the polynomial \( f(x) \).

Short Answer Type Questions - I

Question. If \(p\) and \(q\) are the zeroes of polynomial \(f(x) = 2x^2 - 7x + 3\), find the value of \(p^2 + q^2\).
Answer: We have \(f(x) = 2x^2 - 7x + 3\)
Sum of zeroes \(p + q = -\frac{b}{a} = -\left(\frac{-7}{2}\right) = \frac{7}{2}\)
Product of zeroes \(pq = \frac{c}{a} = \frac{3}{2}\)
Since, \((p + q)^2 = p^2 + q^2 + 2pq\)
so, \(p^2 + q^2 = (p + q)^2 - 2pq\)
\(= \left(\frac{7}{2}\right)^2 - 2\left(\frac{3}{2}\right) = \frac{49}{4} - 3 = \frac{37}{4}\)
Hence \(p^2 + q^2 = \frac{37}{4}\).

Question. Find the condition that zeroes of polynomial \(p(x) = ax^2 + bx + c\) are reciprocal of each other.
Answer: We have \(p(x) = ax^2 + bx + c\)
Let \(\alpha\) and \(\frac{1}{\alpha}\) be the zeroes of \(p(x)\), then
Product of zeroes, \(\frac{c}{a} = \alpha \times \frac{1}{\alpha} = 1\) or \(\frac{c}{a} = 1\)
So, required condition is, \(c = a\)

Question. Find the value of \(k\) if -1 is a zero of the polynomial \(p(x) = kx^2 - 4x + k\).
Answer: We have \(p(x) = kx^2 - 4x + k\)
Since, -1 is a zero of the polynomial, then
\(p(-1) = 0\)
\(k(-1)^2 - 4(-1) + k = 0\)
or, \(k + 4 + k = 0\)
or, \(2k + 4 = 0\)
or, \(2k = -4\)
Hence, \(k = -2\)

Question. If \(\alpha\) and \(\beta\) are the zeroes of a polynomial \(x^2 - 4\sqrt{3}x + 3\), then find the value of \(\alpha + \beta - \alpha\beta\).
Answer: We have \(p(x) = x^2 - 4\sqrt{3}x + 3\)
If \(\alpha\) and \(\beta\) are the zeroes of \(x^2 - 4\sqrt{3}x + 3\), then
Sum of zeroes, \(\alpha + \beta = -\frac{b}{a} = -\frac{(-4\sqrt{3})}{1} = 4\sqrt{3}\)
Product of zeroes \(\alpha\beta = \frac{c}{a} = \frac{3}{1} = 3\)
Now \(\alpha + \beta - \alpha\beta = 4\sqrt{3} - 3\).

Question. Find the values of \(a\) and \(b\), if they are the zeroes of polynomial \(x^2 + ax + b\).
Answer: We have \(p(x) = x^2 + ax + b\)
Since \(a\) and \(b\) are the zeroes of polynomial, we get,
Product of zeroes, \(ab = b \Rightarrow a = 1\)
Sum of zeroes, \(a + b = -a \Rightarrow b = -2a = -2(1) = -2\)

Question. If \(\alpha\) and \(\beta\) are the zeroes of the polynomial \(f(x) = x^2 - 6x + k\), find the value of \(k\), such that \(\alpha^2 + \beta^2 = 40\).
Answer: We have \(f(x) = x^2 - 6x + k\)
Sum of zeroes, \(\alpha + \beta = -\frac{b}{a} = -\frac{(-6)}{1} = 6\)
Product of zeroes, \(\alpha\beta = \frac{c}{a} = \frac{k}{1} = k\)
Now \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 40\)
\((6)^2 - 2k = 40\)
\(36 - 2k = 40\)
\(-2k = 4\)
Thus \(k = -2\)

Question. If one of the zeroes of the quadratic polynomial \(f(x) = 14x^2 - 42k^2x - 9\) is negative of the other, find the value of '\(k\)'.
Answer: We have \(f(x) = 14x^2 - 42k^2x - 9\)
Let one zero be \(\alpha\), then other zero will be \(-\alpha\).
Sum of zeroes \(= \alpha + (-\alpha) = 0\).
Thus sum of zero will be 0.
Sum of zeroes \(0 = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2}\)
\(0 = -\frac{-42k^2}{14} = 3k^2\)
Thus \(k = 0\).

Question. If one zero of the polynomial \(2x^2 + 3x + \lambda\) is \(\frac{1}{2}\), find the value of \(\lambda\) and the other zero.
Answer: Let, the zeroes of \(2x^2 + 3x + \lambda\) be \(\frac{1}{2}\) and \(\beta\).
Product of zeroes \(\frac{1}{2}\beta = \frac{\lambda}{2}\)
or, \(\beta = \lambda\)
and sum of zeroes, \(\frac{1}{2} + \beta = -\frac{3}{2}\)
or \(\beta = -\frac{3}{2} - \frac{1}{2} = -2\)
Hence \(\lambda = \beta = -2\)
Thus other zero is -2.

Question. If \(\alpha\) and \(\beta\) are zeroes of the polynomial \(f(x) = x^2 - x - k\), such that \(\alpha - \beta = 9\), find \(k\).
Answer: We have \(f(x) = x^2 - x - k\)
Since \(\alpha\) and \(\beta\) are the zeroes of the polynomial, then
Sum of zeroes, \(\alpha + \beta = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} = -\left(\frac{-1}{1}\right) = 1\)
\(\alpha + \beta = 1\) ...(i)
Given \(\alpha - \beta = 9\) ...(ii)
Solving (i) and (ii) we get \(\alpha = 5\) and \(\beta = -4\)
Product of zeroes \(\alpha\beta = \frac{\text{Constant term}}{\text{Coefficient of } x^2}\)
or \(\alpha\beta = -k\)
Substituting \(\alpha = 5\) and \(\beta = -4\) we have
\((5)(-4) = -k\)
Thus \(k = 20\)

Question. If the zeroes of the polynomial \(x^2 + px + q\) are double in value to the zeroes of \(2x^2 - 5x - 3\), find the value of \(p\) and \(q\).
Answer: We have \(f(x) = 2x^2 - 5x - 3\)
Let the zeroes of polynomial be \(\alpha\) and \(\beta\), then
Sum of zeroes \(\alpha + \beta = \frac{5}{2}\)
Product of zeroes \(\alpha\beta = -\frac{3}{2}\)
According to the question, zeroes of \(x^2 + px + q\) are \(2\alpha\) and \(2\beta\).
Sum of zeros, \(2\alpha + 2\beta = -\frac{p}{1}\)
\(2(\alpha + \beta) = -p\)
Substituting \(\alpha + \beta = \frac{5}{2}\) we have \(2 \times \frac{5}{2} = -p \Rightarrow p = -5\)
Product of zeroes, \(2\alpha \times 2\beta = \frac{q}{1}\)
\(4\alpha\beta = q\)
Substituting \(\alpha\beta = -\frac{3}{2}\) we have \(4 \times -\frac{3}{2} = q \Rightarrow -6 = q\)
Thus \(p = -5\) and \(q = -6\).

Question. If \(\alpha\) and \(\beta\) are zeroes of \(x^2 - (k-6)x + 2(2k-1)\), find the value of \(k\) if \(\alpha + \beta = \frac{1}{2}\alpha\beta\).
Answer: We have \(p(x) = x^2 - (k-6)x + 2(2k-1)\)
Since \(\alpha, \beta\) are the zeroes of polynomial \(p(x)\), we get
\(\alpha + \beta = -[-(k - 6)] = k - 6\)
\(\alpha\beta = 2(2k - 1)\)
Now \(\alpha + \beta = \frac{1}{2}\alpha\beta\)
Thus \(k + 6 = \frac{2(2k - 1)}{2}\)
or, \(k - 6 = 2k - 1\)
\(k = -5\)
Hence the value of \(k\) is -5.

Short Answer Type Questions - II

Question. Show that \( \frac{1}{2} \) and \( -\frac{3}{2} \) are the zeroes of the polynomial \( 4x^2 + 4x - 3 \) and verify relationship between zeroes and coefficients of the polynomial.
Answer: We have \( p(x) = 4x^2 + 4x - 3 \). If \( \frac{1}{2} \) and \( -\frac{3}{2} \) are the zeroes of the polynomial \( p(x) \), then these must satisfy \( p(x) = 0 \). \( p\left(\frac{1}{2}\right) = 4\left(\frac{1}{4}\right) + 4\left(\frac{1}{2}\right) - 3 = 1 + 2 - 3 = 0 \). And \( p\left(-\frac{3}{2}\right) = 4\left(\frac{9}{4}\right) + 4\left(-\frac{3}{2}\right) - 3 = 9 - 6 - 3 = 0 \). Thus, \( \frac{1}{2}, -\frac{3}{2} \) are zeroes of polynomial \( 4x^2 + 4x - 3 \). Sum of zeroes \( = \frac{1}{2} - \frac{3}{2} = -1 = -\frac{4}{4} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} \). Product of zeroes \( = \left(\frac{1}{2}\right)\left(-\frac{3}{2}\right) = -\frac{3}{4} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} \). Verified.

Question. Find the zeroes of the quadratic polynomial \( x^2 - 2\sqrt{2}x \) and verify the relationship between the zeroes and the coefficients.
Answer: We have \( p(x) = x^2 - 2\sqrt{2}x = 0 \), \( x(x - 2\sqrt{2}) = 0 \). Thus zeroes are \( 0 \) and \( 2\sqrt{2} \). Sum of zeroes \( 2\sqrt{2} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} \) and product of zeroes \( 0 = \frac{\text{Constant term}}{\text{Coefficient of } x^2} \). Hence verified.

Question. Find the zeroes of the quadratic polynomial \( 5x^2 + 8x - 4 \) and verify the relationship between the zeroes and the coefficients of the polynomial.
Answer: We have \( p(x) = 5x^2 + 8x - 4 = 0 \), \( 5x^2 + 10x - 2x - 4 = 0 \), \( 5x(x+2) - 2(x+2) = 0 \), \( (x+2)(5x-2) = 0 \). Substituting \( p(x) = 0 \) we get zeroes as \( -2 \) and \( \frac{2}{5} \). Verification: Sum of zeroes \( = -2 + \frac{2}{5} = \frac{-8}{5} \). Product of zeroes \( = (-2) \times \left(\frac{2}{5}\right) = \frac{-4}{5} \). Now from polynomial we have Sum of zeroes \( = -\frac{b}{a} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} = -\frac{8}{5} \). Product of zeroes \( = \frac{c}{a} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} = -\frac{4}{5} \). Hence Verified.

Question. If \( \alpha \) and \( \beta \) are the zeroes of a quadratic polynomial such that \( \alpha + \beta = 24 \) and \( \alpha - \beta = 8 \). Find the quadratic polynomial having \( \alpha \) and \( \beta \) as its zeroes.
Answer: We have \( \alpha + \beta = 24 \) ...(1), \( \alpha - \beta = 8 \) ...(2). Adding equations (1) and (2) we have \( 2\alpha = 32 \Rightarrow \alpha = 16 \). Subtracting (2) from (1) we have \( 2\beta = 16 \Rightarrow \beta = 8 \). (Note: The printed text has a typo, \( 24 - 8 = 16 \)). Wait, looking at the step: \( \alpha + \beta = 24 \), \( \alpha - \beta = 8 \), adding gives \( 2\alpha = 32 \rightarrow \alpha = 16 \). Subtracting gives \( 2\beta = 16 \rightarrow \beta = 8 \). The image shows \( 2\beta = 24 \) which is an error in the original solution text, but the final steps use \( \alpha=16 \) and \( \beta=8 \). Hence, the quadratic polynomial \( p(x) = x^2 - (\alpha + \beta)x + \alpha\beta = x^2 - (16 + 8)x + (16)(8) = x^2 - 24x + 128 \).

Question. If \( \alpha, \beta, \) and \( \gamma \) are zeroes of the polynomial \( 6x^3 + 3x^2 - 5x + 1 \), then find the value of \( \alpha^{-1} + \beta^{-1} + \gamma^{-1} \).
Answer: We have \( p(x) = 6x^3 + 3x^2 - 5x + 1 \). Since \( \alpha, \beta, \) and \( \gamma \) are zeroes polynomial \( p(x) \), we have \( \alpha + \beta + \gamma = -\frac{b}{a} = -\frac{3}{6} = -\frac{1}{2} \). \( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = -\frac{5}{6} \). And \( \alpha\beta\gamma = -\frac{d}{a} = -\frac{1}{6} \). Now \( \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma} = \frac{-5/6}{-1/6} = \frac{-5}{6} \times \frac{6}{-1} = 5 \). Hence \( \alpha^{-1} + \beta^{-1} + \gamma^{-1} = 5 \).

LONG ANSWER TYPE QUESTIONS

Question. Find the value of \( a \) and \( b \) so that \( 8x^4 + 14x^3 - 2x^2 + ax + b \) is exactly divisible by \( 4x^2 + 3x - 2 \).
Answer: By long division, the remainder is obtained as \( (a + 7)x + b - 2 \).
For exact division, remainder is zero, so
\( (a + 7)x + b - 2 = 0 \)
or \( a + 7 = 0 \), \( b - 2 = 0 \)
\( \Rightarrow a = -7, b = 2 \)

Question. On dividing a polynomial \( 3x^3 + 4x^2 + 5x - 13 \) by a polynomial \( g(x) \), the quotient and the remainder are \( (3x + 10) \) and \( (16x - 43) \) respectively. Find \( g(x) \).
Answer: \( 3x^3 + 4x^2 + 5x - 13 = (3x + 10)g(x) + (16x - 43) \)
\( g(x)(3x + 10) = (3x^3 + 4x^2 + 5x - 13) - (16x - 43) \)
\( g(x)(3x + 10) = 3x^3 + 4x^2 - 11x + 30 \)
Hence, \( g(x) = x^2 - 2x + 3 \)

Question. When \( p(x) = x^2 + 7x + 9 \) is divisible by \( g(x) \), we get \( (x + 2) \) and \( -1 \) as the quotient and remainder respectively, find \( g(x) \).
Answer: We have \( p(x) = x^2 + 7x + 9 \)
\( q(x) = x + 2 \)
\( r(x) = -1 \)
Now \( p(x) = g(x)q(x) + r(x) \)
\( x^2 + 7x + 9 = g(x)(x + 2) - 1 \)
or, \( g(x) = \frac{x^2 + 7x + 10}{x + 2} \)
\( = \frac{(x + 2)(x + 5)}{(x + 2)} = x + 5 \)
Thus \( g(x) = x + 5 \)

Question. Check by divisible algorithm whether \( x^2 - 2 \) is a factor of \( x^4 + x^3 + x^2 - 2x - 3 \).
Answer: By performing division of \( x^4 + x^3 + x^2 - 2x - 3 \) by \( x^2 - 2 \):
Remainder is found to be \( 3 \).
Since Remainder \( \neq 0 \), hence \( x^2 - 2 \) is not a factor of the given polynomial.

Question. On dividing \( x^4 - x^3 - 3x^2 + 3x + 2 \) by a polynomial \( g(x) \), the quotient and the remainder are \( x^2 - x - 2 \) and \( 2x \) respectively. Find \( g(x) \).
Answer: \( x^4 - x^3 - 3x^2 + 3x + 2 = (x^2 - x - 2)g(x) + 2x \)
\( g(x)(x^2 - x - 2) = (x^4 - x^3 - 3x^2 + 3x + 2) - 2x \)
\( g(x) = \frac{x^4 - x^3 - 3x^2 + x + 2}{x^2 - x - 2} \)
Hence, \( g(x) = x^2 - 1 \)

Question. What should be added in the polynomial \( x^3 + 2x^2 - 9x + 1 \) so that it is completely divisible by \( x + 4 \).
Answer: Let \( k \) be added.
Dividing \( x^3 + 2x^2 - 9x + 1 \) by \( x + 4 \), the remainder is \( 5 \).
Remainder should be zero
\( 5 + k = 0 \)
Hence –5 should be added.

Question. If the polynomial \( f(x) = 3x^4 + 3x^3 - 11x^2 - 5x + 10 \) is completely divisible by \( 3x^2 - 5 \), find all its zeroes.
Answer: Since \( 3x^2 - 5 \) divides \( f(x) \) completely, \( (3x^2 - 5) = 0 \Rightarrow x = \pm \sqrt{\frac{5}{3}} \).
Dividing \( f(x) \) by \( 3x^2 - 5 \), we get quotient \( x^2 + x - 2 \).
Since \( (x^2 + x - 2) \) is a factor of \( p(x) \).
\( x^2 + x - 2 = 0 \)
Factorising it, we get \( x = -2 \) and \( 1 \)
Thus -2 and 1 are zeroes or \( p(x) \).
All the zeroes of \( p(x) \) are \( \sqrt{\frac{5}{3}}, -\sqrt{\frac{5}{3}}, -2 \) and \( 1 \).

Long Answer Type Questions

Question. Show that 3 is a zero of the polynomial \( 2x^3 - x^2 - 13x - 6 \). Hence find all the zeroes of this polynomial.
Answer: \( p(3) = 2(3)^3 - (3)^2 - 13(3) - 6 = 54 - 9 - 39 - 6 = 0 \).
So, \( x - 3 \) is a factor. Dividing \( p(x) \) by \( x - 3 \), quotient is \( 2x^2 + 5x + 2 \).
Factorising quotient: \( 2x^2 + 4x + x + 2 = (2x + 1)(x + 2) \).
Zeroes are \( x = -\frac{1}{2}, -2 \).
All zeroes are \( -\frac{1}{2}, -2, 3 \).

Question. Obtain all other zeroes of the polynomial \( x^4 + 6x^3 + x^2 - 24x - 20 \), if two of its zeroes are +2 and -5.
Answer: Product of factors \( = (x - 2)(x + 5) = x^2 + 3x - 10 \).
Dividing \( x^4 + 6x^3 + x^2 - 24x - 20 \) by \( x^2 + 3x - 10 \), quotient is \( x^2 + 3x + 2 \).
Factorising: \( x^2 + 3x + 2 = (x + 2)(x + 1) \).
Hence other two zeroes are -2 and -1.

Question. Obtain all other zeroes of the polynomial \( 4x^4 + x^3 - 72x^2 - 18x \), if two of its zeroes are \( 3\sqrt{2} \) and \( -3\sqrt{2} \).
Answer: Product of factors \( = (x - 3\sqrt{2})(x + 3\sqrt{2}) = x^2 - 18 \).
Dividing \( 4x^4 + x^3 - 72x^2 - 18x \) by \( x^2 - 18 \), quotient is \( 4x^2 + x \).
\( 4x^2 + x = x(4x + 1) \).
Zeroes are \( x = 0 \) and \( x = -\frac{1}{4} \).
Hence, other two zeroes are 0 and \( -\frac{1}{4} \).

Question. Obtain all other zeroes of the polynomial \( 9x^4 - 6x^3 - 35x^2 + 24x - 4 \), if two of its zeroes are 2 and -2.
Answer: Product of factors \( = (x - 2)(x + 2) = x^2 - 4 \).
Dividing \( 9x^4 - 6x^3 - 35x^2 + 24x - 4 \) by \( x^2 - 4 \), quotient is \( 9x^2 - 6x + 1 \).
\( 9x^2 - 6x + 1 = (3x - 1)^2 \).
Hence, other two zeroes are \( \frac{1}{3}, \frac{1}{3} \).

Question. Find all the zeros of the polynomial \( 3x^4 + 6x^3 - 2x^2 - 10x - 5 \) if two of its zeroes are \( \sqrt{\frac{5}{3}} \) and \( -\sqrt{\frac{5}{3}} \).
Answer: [Sample Paper 2017]
Product of factors \( = (x - \sqrt{5/3})(x + \sqrt{5/3}) = x^2 - 5/3 \).
Multiplying by 3, we use factor \( 3x^2 - 5 \).
Dividing \( 3x^4 + 6x^3 - 2x^2 - 10x - 5 \) by \( 3x^2 - 5 \), quotient is \( x^2 + 2x + 1 \).
\( x^2 + 2x + 1 = (x + 1)^2 \).
Other zeroes are -1 and -1.
All zeroes: \( \sqrt{5/3}, -\sqrt{5/3}, -1, -1 \).

Hots Questions

Question. Find the value for \( k \) for which \( x^4 + 10x^3 + 25x^2 + 15x + k \) is exactly divisible by \( x + 7 \).
Answer: \( f(x) = x^4 + 10x^3 + 25x^2 + 15x + k \)
Substituting \( x = -7 \):
\( (-7)^4 + 10(-7)^3 + 25(-7)^2 + 15(-7) + k = 0 \)
\( 2401 - 3430 + 1225 - 105 + k = 0 \)
\( 91 + k = 0 \Rightarrow k = -91 \)

Question. If two zeroes of the polynomial \( p(x) = x^4 - 6x^3 - 26x^2 + 138x - 35 \) are \( 2 \pm \sqrt{3} \). Find the other zeroes.
Answer: Product of zeroes \( = [x - (2 + \sqrt{3})][x - (2 - \sqrt{3})] = (x - 2)^2 - (\sqrt{3})^2 = x^2 - 4x + 1 \).
Dividing \( p(x) \) by \( x^2 - 4x + 1 \), quotient is \( x^2 - 2x - 35 \).
Factorising \( x^2 - 2x - 35 = (x + 5)(x - 7) \).
Hence, other two zeroes of \( p(x) \) are -5 and 7.