CBSE Class 10 Maths HOTs Polynomials Set B

Very Short Answer Type Questions

Question. If \(\alpha\) and \(\beta\) are the roots of \(ax^2 - bx + c = 0 (a \neq 0)\), then calculate \(\alpha + \beta\).
Answer: We know that
Sum of the roots \(= -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}\)
Thus \(\alpha + \beta = -\left(\frac{-b}{a}\right) = \frac{b}{a}\)

Question. Calculate the zeroes of the polynomial \(p(x) = 4x^2 - 12x + 9\).
Answer: We have \(p(x) = 4x^2 - 12x + 9\)
\(= 4x^2 - 6x - 6x + 9\)
\(= 2x(2x - 3) - 3(2x - 3)\)
\(= (2x - 3)(2x - 3)\)
Substituting \(p(x) = 0\), and solving we get \(x = \frac{3}{2}, \frac{3}{2}\).
Hence, zeroes of the polynomial are \(\frac{3}{2}, \frac{3}{2}\).

Question. What should be added in the polynomial \( x^3 - 6x^2 + 11x + 8 \) so that it is completely divisible by \( x^2 - 3x + 2 \)?
Answer: Dividing \( x^3 - 6x^2 + 11x + 8 \) by \( x^2 - 3x + 2 \), we get a remainder of \( 14 \). Since Remainder \( = 14 \), to make it \( 0 \), \( -14 \) should be added.

Question. If the polynomial \( 6x^4 + 8x^3 + 17x^2 + 21x + 7 \) is divided by another polynomial \( 3x^2 + 4x + 1 \), the remainder comes out to be \( (ax + b) \), find the values of \( a \) and \( b \).
Answer: Dividing \( 6x^4 + 8x^3 + 17x^2 + 21x + 7 \) by \( 3x^2 + 4x + 1 \), the long division yields a quotient of \( 2x^2 + 5 \) and a remainder of \( x + 2 \). Comparing \( x + 2 \) with \( ax + b \), we get \( a = 1 \) and \( b = 2 \).

Short Answer Type Questions - I

Question. If \( x^3 - 6x^2 + 6x + k \) is completely divisible by \( x - 3 \), then find the value of \( k \).
Answer: By performing long division of \( x^3 - 6x^2 + 6x + k \) by \( x - 3 \):
Remainder = \( k - 9 \)
Remainder should be zero
\( k - 9 = 0 \)
So, \( k = 9 \)

Question. Divide the polynomial \( p(x) = x^3 - 4x + 6 \) by the polynomial \( g(x) = 2 - x^2 \) and find the quotient and the remainder.
Answer: By dividing \( x^3 - 4x + 6 \) by \( -x^2 + 2 \):
Thus, Quotient = \( -x \)
and Remainder = \( 6 - 2x \)

Question. Divide the polynomial \( p(x) = x^2 - 5x + 16 \) by the polynomial \( g(x) = x - 2 \) and find the quotient and the remainder.
Answer: By performing long division:
Quotient = \( x - 3 \), Remainder = \( 10 \

Question. If zeroes of the polynomial \(x^2 + 4x + 2a\) are \(\alpha\) and \(\frac{2}{\alpha}\), then find the value of \(a\).
Answer: Product of (zeroes) roots,
\(\frac{c}{a} = \frac{2a}{1} = \alpha \times \frac{2}{\alpha} = 2\)
or, \(2a = 2\)
Thus \(a = 1\)

Question. Find all the zeroes of \(f(x) = x^2 - 2x\).
Answer: We have \(f(x) = x^2 - 2x = x(x - 2)\)
Substituting \(f(x) = 0\), and solving we get \(x = 0, 2\).
Hence, zeroes are 0 and 2.

Question. Find the zeroes of the quadratic polynomial \(\sqrt{3}x^2 - 8x + 4\sqrt{3}\).
Answer: \(p(x) = \sqrt{3}x^2 - 8x + 4\sqrt{3} = 0\)
\(= \sqrt{3}x^2 - 6x - 2x + 4\sqrt{3} = 0\)
\(= \sqrt{3}x(x - 2\sqrt{3}) - 2(x - 2\sqrt{3}) = 0\)
\(= (\sqrt{3}x - 2)(x - 2\sqrt{3}) = 0\)
Substituting \(p(x) = 0\), and solving we get \(x = \frac{2}{\sqrt{3}}, 2\sqrt{3}\).
Hence, zeroes are \(\frac{2}{\sqrt{3}}\) and \(2\sqrt{3}\).

Question. Find a quadratic polynomial, the sum and product of whose zeroes are 6 and 9 respectively. Hence find the zeroes.
Answer: Sum of zeroes, \(\alpha + \beta = 6\)
Product of zeroes \(\alpha\beta = 9\)
Now \(p(x) = x^2 - (\alpha + \beta)x + \alpha\beta\)
Thus \(x^2 - 6x + 9\)
Thus quadratic polynomial is \(x^2 - 6x + 9\).
Now \(p(x) = x^2 - 6x + 9 = (x - 3)(x - 3)\)
Substituting \(p(x) = 0\), we get \(x = 3, 3\),
Hence zeroes are 3, 3.

Question. Find the quadratic polynomial whose sum and product of the zeroes are \(\frac{21}{8}\) and \(\frac{5}{16}\) respectively.
Answer: Sum of zeroes, \(\alpha + \beta = \frac{21}{8}\)
Product of zeroes \(\alpha\beta = \frac{5}{16}\)
Now \(p(x) = x^2 - (\alpha + \beta)x + \alpha\beta\)
\(= x^2 - \frac{21}{8}x + \frac{5}{16}\)
or \(p(x) = \frac{1}{16}(16x^2 - 42x + 5)\)

Question. Form a quadratic polynomial \(p(x)\) with 3 and \(-\frac{2}{5}\) as sum and product of its zeroes, respectively.
Answer: Sum of zeroes, \(\alpha + \beta = 3\)
Product of zeroes \(\alpha\beta = -\frac{2}{5}\)
Now \(p(x) = x^2 - (\alpha + \beta)x + \alpha\beta\)
\(= x^2 - 3x - \frac{2}{5}\)
\(= \frac{1}{5}(5x^2 - 15x - 2)\)
The required quadratic polynomial is \(\frac{1}{5}(5x^2 - 15x - 2)\).

Question. What should added to the polynomial \(x^3 - 3x^2 + 6x - 15\) so that it is completely divisible by \(x - 3\).
Answer: We divide \(x^3 - 3x^2 + 6x - 15\) by \(x - 3\) as follows.
\(x - 3 \overline{ | x^3 - 3x^2 + 6x - 15 }\) (Quotient: \(x^2 + 6\))
\(x^3 - 3x^2\)
--------------
\(6x - 15\)
\(6x - 18\)
--------------
         \(3\)
Here remainder is 3, hence -3 must be added so that there is no remainder.

Question. If \(m\) and \(n\) are the zeroes of the polynomial \(3x^2 + 11x - 4\), find the value of \(\frac{m}{n} + \frac{n}{m}\).
Answer: We have \(\frac{m}{n} + \frac{n}{m} = \frac{m^2 + n^2}{mn} = \frac{(m + n)^2 - 2mn}{mn}\) --- (1)
Sum of zeroes \(m + n = -\frac{11}{3}\)
Product of zeroes \(mn = -\frac{4}{3}\)
Substituting in (1) we have
\(\frac{m}{n} + \frac{n}{m} = \frac{(-\frac{11}{3})^2 - 2(-\frac{4}{3})}{-\frac{4}{3}} = \frac{\frac{121}{9} + \frac{8}{3}}{-\frac{4}{3}} = \frac{121 + 24}{-12}\)
or \(\frac{m}{n} + \frac{n}{m} = -\frac{145}{12}\)

Short Answer Type Questions - II

Question. Verify whether 2, 3 and \(\frac{1}{2}\) are the zeroes of the polynomial \(p(x) = 2x^3 - 11x^2 + 17x - 6\).
Answer: If 2, 3 and \(\frac{1}{2}\) are the zeroes of the polynomial \(p(x)\), then these must satisfy \(p(x) = 0\)
(1) 2, \(p(2) = 2(2)^3 - 11(2)^2 + 17(2) - 6 = 16 - 44 + 34 - 6 = 50 - 50 = 0\)
(2) 3, \(p(3) = 2(3)^3 - 11(3)^2 + 17(3) - 6 = 54 - 99 + 51 - 6 = 105 - 105 = 0\)
(3) \(\frac{1}{2}\), \(p\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 - 11\left(\frac{1}{2}\right)^2 + 17\left(\frac{1}{2}\right) - 6 = \frac{1}{4} - \frac{11}{4} + \frac{17}{2} - 6 = 0\)
Hence, 2, 3, and \(\frac{1}{2}\) are the zeroes of \(p(x)\).

Question. If the sum and product of the zeroes of the polynomial \(ax^2 - 5x + c\) are equal to 10 each, find the value of '\(a\)' and '\(c\)'.
Answer: We have \(f(x) = ax^2 - 5x + c\)
Let the zeroes of \(f(x)\) be \(\alpha\) and \(\beta\), then,
Sum of zeroes \(\alpha + \beta = -\frac{-5}{a} = \frac{5}{a}\)
Product of zeroes \(\alpha\beta = \frac{c}{a}\)
According to question, the sum and product of the zeroes of the polynomial \(f(x)\) are equal to 10 each.
Thus \(\frac{5}{a} = 10 \Rightarrow a = \frac{1}{2}\) ...(1)
and \(\frac{c}{a} = 10 \Rightarrow c = 10a\) ...(2)
Substituting \(a = \frac{1}{2}\) in (2) we get \(c = 10(\frac{1}{2}) = 5\)
Hence \(a = \frac{1}{2}\) and \(c = 5\).

Question. If one the zero of a polynomial \(3x^2 - 8x + 2k + 1\) is seven times the other, find the value of \(k\).
Answer: We have \(f(x) = 3x^2 - 8x + 2k + 1\)
Let \(\alpha\) and \(\beta\) be the zeroes of the polynomial, then \(\beta = 7\alpha\)
Sum of zeroes, \(\alpha + \beta = -\left(\frac{-8}{3}\right)\)
\(\alpha + 7\alpha = 8\alpha = \frac{8}{3}\)
So \(\alpha = \frac{1}{3}\)
Product of zeroes, \(\alpha \times 7\alpha = \frac{2k + 1}{3}\)
\(7\alpha^2 = \frac{2k + 1}{3}\)

Question. Quadratic polynomial \( 2x^2 - 3x + 1 \) has zeroes as \( \alpha \) and \( \beta \). Now form a quadratic polynomial whose zeroes are \( 3\alpha \) and \( 3\beta \).
Answer: We have \( f(x) = 2x^2 - 3x + 1 \). If \( \alpha \) and \( \beta \) are the zeroes of \( 2x^2 - 3x + 1 \), then Sum of zeroes \( \alpha + \beta = \frac{-b}{a} = \frac{3}{2} \). Product of zeroes \( \alpha\beta = \frac{c}{a} = \frac{1}{2} \). New quadratic polynomial whose zeroes are \( 3\alpha \) and \( 3\beta \) is, \( p(x) = x^2 - (3\alpha + 3\beta)x + 3\alpha \times 3\beta = x^2 - 3(\alpha + \beta)x + 9\alpha\beta = x^2 - 3\left(\frac{3}{2}\right)x + 9\left(\frac{1}{2}\right) = x^2 - \frac{9}{2}x + \frac{9}{2} = \frac{1}{2}(2x^2 - 9x + 9) \). Hence, required quadratic polynomial is \( \frac{1}{2}(2x^2 - 9x + 9) \).

Question. If \( \alpha \) and \( \beta \) are the zeroes of the polynomial \( 6y^2 - 7y + 2 \), find a quadratic polynomial whose zeroes are \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \).
Answer: We have \( p(y) = 6y^2 - 7y + 2 \). Sum of zeroes \( \alpha + \beta = -\left(-\frac{7}{6}\right) = \frac{7}{6} \). Product of zeroes \( \alpha\beta = \frac{2}{6} = \frac{1}{3} \). Sum of zeroes of new polynomial \( g(y) \): \( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{7/6}{1/3} = \frac{7}{6} \times \frac{3}{1} = \frac{7}{2} \). And product of zeroes of new polynomial \( g(y) \): \( \frac{1}{\alpha} \times \frac{1}{\beta} = \frac{1}{\alpha\beta} = \frac{1}{1/3} = 3 \). The required polynomial is \( g(x) = y^2 - \frac{7}{2}y + 3 = \frac{1}{2}[2y^2 - 7y + 6] \).

LONG ANSWER TYPE QUESTIONS

Question. Polynomial \( x^4 + 7x^3 + 7x^2 + px + q \) is exactly divisible by \( x^2 + 7x + 12 \), then find the value of \( p \) and \( q \).
Answer: We have \( f(x) = x^4 + 7x^3 + 7x^2 + px + q \). Now \( x^2 + 7x + 12 = 0 \Rightarrow x^2 + 4x + 3x + 12 = 0 \Rightarrow x(x+4) + 3(x+4) = 0 \Rightarrow (x+4)(x+3) = 0 \Rightarrow x = -4, -3 \). Since \( f(x) \) is exactly divisible by \( x^2 + 7x + 12 \), then \( x = -4 \) and \( x = -3 \) must be its zeroes and these must satisfy \( f(x) = 0 \). So putting \( x = -4 \) and \( x = -3 \) in \( f(x) \) and equating to zero we get \( f(-4) : (-4)^4 + 7(-4)^3 + 7(-4)^2 + p(-4) + q = 0 \Rightarrow 256 - 448 + 112 - 4p + q = 0 \Rightarrow -4p + q - 80 = 0 \Rightarrow 4p - q = -80 \)...(1). \( f(-3) : (-3)^4 + 7(-3)^3 + 7(-3)^2 + p(-3) + q = 0 \Rightarrow 81 - 189 + 63 - 3p + q = 0 \Rightarrow -3p + q - 45 = 0 \Rightarrow 3p - q = -45 \)...(2). Subtracting eq. (2) from (1) we have \( p = -35 \). On putting the value of \( p \) in eq. (1) we have \( 4(-35) - q = -80 \Rightarrow -140 - q = -80 \Rightarrow -q = 140 - 80 \Rightarrow -q = 60 \Rightarrow q = -60 \). Hence, \( p = -35 \) and \( q = -60 \).

Question. If \( \alpha \) and \( \beta \) are the zeroes of the polynomial \( p(x) = 2x^2 + 5x + k \) satisfying the relation, \( \alpha^2 + \beta^2 + \alpha\beta = \frac{21}{4} \), then find the value of \( k \).
Answer: We have \( p(x) = 2x^2 + 5x + k \). Sum of zeroes, \( \alpha + \beta = -\frac{b}{a} = -\frac{5}{2} \). Product of zeroes, \( \alpha\beta = \frac{c}{a} = \frac{k}{2} \). According to the question, \( \alpha^2 + \beta^2 + \alpha\beta = \frac{21}{4} \Rightarrow \alpha^2 + \beta^2 + 2\alpha\beta - \alpha\beta = \frac{21}{4} \Rightarrow (\alpha + \beta)^2 - \alpha\beta = \frac{21}{4} \). Substituting values we have \( \left(-\frac{5}{2}\right)^2 - \frac{k}{2} = \frac{21}{4} \Rightarrow \frac{25}{4} - \frac{k}{2} = \frac{21}{4} \Rightarrow \frac{k}{2} = \frac{25}{4} - \frac{21}{4} = \frac{4}{4} = 1 \Rightarrow k = 2 \).

Question. If \( \alpha \) and \( \beta \) are the zeroes of polynomial \( p(x) = 3x^2 + 2x + 1 \), find the polynomial whose zeroes are \( \frac{1-\alpha}{1+\alpha} \) and \( \frac{1-\beta}{1+\beta} \).
Answer: We have \( p(x) = 3x^2 + 2x + 1 \). Since \( \alpha \) and \( \beta \) are zeroes, \( \alpha + \beta = -\frac{2}{3} \) and \( \alpha\beta = \frac{1}{3} \). Let \( \alpha_1 \) and \( \beta_1 \) be zeros of new polynomial \( q(x) \). Sum of zeroes, \( \alpha_1 + \beta_1 = \frac{1-\alpha}{1+\alpha} + \frac{1-\beta}{1+\beta} = \frac{(1-\alpha)(1+\beta) + (1+\alpha)(1-\beta)}{(1+\alpha)(1+\beta)} = \frac{1+\beta-\alpha-\alpha\beta+1-\beta+\alpha-\alpha\beta}{1+\alpha+\beta+\alpha\beta} = \frac{2-2\alpha\beta}{1+(\alpha+\beta)+\alpha\beta} = \frac{2-2(1/3)}{1+(-2/3)+1/3} = \frac{4/3}{2/3} = 2 \). Product of zeroes, \( \alpha_1\beta_1 = \left(\frac{1-\alpha}{1+\alpha}\right)\left(\frac{1-\beta}{1+\beta}\right) = \frac{1-\beta-\alpha+\alpha\beta}{1+\alpha+\beta+\alpha\beta} = \frac{1-(\alpha+\beta)+\alpha\beta}{1+(\alpha+\beta)+\alpha\beta} = \frac{1-(-2/3)+1/3}{1+(-2/3)+1/3} = \frac{6/3}{2/3} = 3 \). Hence, Required polynomial \( q(x) = x^2 - (\alpha_1 + \beta_1)x + \alpha_1\beta_1 = x^2 - 2x + 3 \).

Question. If \( \alpha \) and \( \beta \) are the zeroes of the polynomial \( x^2 + 4x + 3 \), find the polynomial whose zeroes are \( 1 + \frac{\beta}{\alpha} \) and \( 1 + \frac{\alpha}{\beta} \).
Answer: We have \( p(x) = x^2 + 4x + 3 \). Since \( \alpha \) and \( \beta \) are zeroes, \( \alpha + \beta = -4 \) and \( \alpha\beta = 3 \). Let \( \alpha_1 \) and \( \beta_1 \) be zeroes of new polynomial \( q(x) \). Sum of zeroes, \( \alpha_1 + \beta_1 = \left(1 + \frac{\beta}{\alpha}\right) + \left(1 + \frac{\alpha}{\beta}\right) = 2 + \frac{\beta}{\alpha} + \frac{\alpha}{\beta} = 2 + \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{2\alpha\beta + \alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha + \beta)^2}{\alpha\beta} = \frac{(-4)^2}{3} = \frac{16}{3} \). Product of zeroes, \( \alpha_1\beta_1 = \left(1 + \frac{\beta}{\alpha}\right)\left(1 + \frac{\alpha}{\beta}\right) = \left(\frac{\alpha + \beta}{\alpha}\right)\left(\frac{\beta + \alpha}{\beta}\right) = \frac{(\alpha + \beta)^2}{\alpha\beta} = \frac{16}{3} \). Hence, Required polynomial \( q(x) = x^2 - (\alpha_1 + \beta_1)x + \alpha_1\beta_1 = x^2 - \frac{16}{3}x + \frac{16}{3} = \frac{1}{3}(3x^2 - 16x + 16) \).

Question. If \( \alpha \) and \( \beta \) are zeroes of the polynomial \( p(x) = 6x^2 - 5x + k \) such that \( \alpha - \beta = \frac{1}{6} \). Find the value of \( k \).
Answer: We have \( p(x) = 6x^2 - 5x + k \). Sum of zeroes, \( \alpha + \beta = -\left(-\frac{5}{6}\right) = \frac{5}{6} \)...(1). Product of zeroes \( \alpha\beta = \frac{k}{6} \)...(2). Given \( \alpha - \beta = \frac{1}{6} \)...(3). Solving (1) and (3) we get \( 2\alpha = 1 \Rightarrow \alpha = \frac{1}{2} \) and \( \beta = \frac{1}{3} \). Substituting the values in (2) we have \( \frac{1}{2} \times \frac{1}{3} = \frac{k}{6} \Rightarrow \frac{1}{6} = \frac{k}{6} \Rightarrow k = 1 \).

Question. If \( \beta \) and \( \frac{1}{\beta} \) are zeroes of the polynomial \( (a^2 + a)x^2 + 61x + 6a \). Find the value of \( \beta \) and \( a \).
Answer: We have \( p(x) = (a^2 + a)x^2 + 61x + 6a \). Since \( \beta \) and \( \frac{1}{\beta} \) are zeroes, Product of zeroes \( \beta \times \frac{1}{\beta} = \frac{6a}{a^2 + a} \Rightarrow 1 = \frac{6a}{a(a + 1)} \Rightarrow a + 1 = 6 \Rightarrow a = 5 \). Substituting \( a = 5 \) in the sum of zeroes: \( \beta + \frac{1}{\beta} = -\frac{61}{5^2 + 5} = -\frac{61}{30} \Rightarrow \frac{\beta^2 + 1}{\beta} = -\frac{61}{30} \Rightarrow 30\beta^2 + 61\beta + 30 = 0 \). Solving using quadratic formula: \( \beta = \frac{-61 \pm \sqrt{61^2 - 4(30)(30)}}{2(30)} = \frac{-61 \pm 11}{60} \). Thus \( \beta = -\frac{50}{60} = -\frac{5}{6} \) or \( \beta = -\frac{72}{60} = -\frac{6}{5} \). Hence \( a = 5, \beta = -\frac{5}{6}, -\frac{6}{5} \).

Question. What should be added to \( x^3 + 5x^2 + 7x + 3 \) so that it is completely divisible by \( x^2 + 2x \).
Answer: On dividing \( x^3 + 5x^2 + 7x + 3 \) by \( x^2 + 2x \), we get quotient \( x + 3 \) and remainder \( x + 3 \).
To make it completely divisible, we should add the negative of the remainder.
Hence, \( -(x + 3) \) should be added.

Question. Divided \( 6x^3 + 2x^2 - 4x + 3 \) by \( 3x^2 - 2x + 1 \) and verify the division algorithm.
Answer: By long division:
Quotient = \( 2x + 2 \); Remainder = \( -2x + 1 \)
Verification: \( p(x) = g(x)q(x) + r(x) \)
\( = (3x^2 - 2x + 1)(2x + 2) + (-2x + 1) \)
\( = 6x^3 + 6x^2 - 4x^2 - 4x + 2x + 2 - 2x + 1 \)
\( = 6x^3 + 2x^2 - 4x + 3 \) Verified.

Long Answer Type Questions

Question. If two zeroes of a polynomial \( x^3 + 5x^2 + 7x + 3 \) are -1 and -3, then find the third zero.
Answer: Zeroes are \( x = -1 \) and \( x = -3 \). Factors are \( (x+1) \) and \( (x+3) \).
Product of factors \( = (x+1)(x+3) = x^2 + 4x + 3 \).
Dividing \( x^3 + 5x^2 + 7x + 3 \) by \( x^2 + 4x + 3 \), we get quotient \( x + 1 \).
The third zero is obtained from \( x + 1 = 0 \Rightarrow x = -1 \).
The third zero is -1.

Question. Given that \( x - \sqrt{5} \) is a factor of the polynomial \( x^3 - 3\sqrt{5}x^2 - 5x + 15\sqrt{5} \), find all the zeroes of the polynomial.
Answer: Dividing \( x^3 - 3\sqrt{5}x^2 - 5x + 15\sqrt{5} \) by \( x - \sqrt{5} \), we get quotient \( x^2 - 2\sqrt{5}x - 15 \).
Factorising the quotient:
\( x^2 - 2\sqrt{5}x - 15 = x^2 - 3\sqrt{5}x + \sqrt{5}x - 15 \)
\( = x(x - 3\sqrt{5}) + \sqrt{5}(x - 3\sqrt{5}) \)
\( = (x + \sqrt{5})(x - 3\sqrt{5}) \)
All the zeroes are \( \sqrt{5}, -\sqrt{5} \) and \( 3\sqrt{5} \).

Question. If the polynomial \( x^4 - 6x^3 + 16x^2 - 25x + 10 \) is divided by \( (x^2 - 2x + k) \), the remainder comes out to be \( x + a \), find \( k \) and \( a \).
Answer: By long division, remainder is \( (2k - 9)x + (10 - 8k + k^2) \).
Given, remainder \( = x + a \)
Comparing coefficients of \( x \):
\( 2k - 9 = 1 \Rightarrow k = 5 \)
Substituting \( k = 5 \) into the constant part:
\( a = 10 - 8(5) + 5^2 = 10 - 40 + 25 = -5 \)
\( k = 5, a = -5 \)

Question. Find the other zeroes of the polynomial \( x^4 - 5x^3 + 2x^2 + 10x - 8 \) if it is given that two zeroes are \( -\sqrt{2} \) and \( \sqrt{2} \).
Answer: Product of factors \( = (x + \sqrt{2})(x - \sqrt{2}) = x^2 - 2 \).
Dividing \( x^4 - 5x^3 + 2x^2 + 10x - 8 \) by \( x^2 - 2 \), quotient is \( x^2 - 5x + 4 \).
Factorising quotient: \( x^2 - 5x + 4 = (x - 4)(x - 1) \).
Hence other zeroes are 4 and 1.

Hots Questions

Question. If \( \alpha \) and \( \beta \) are the zeroes the polynomial \( 2x^2 - 4x + 5 \), find the values of
(i) \( \alpha^2 + \beta^2 \)
(ii) \( \frac{1}{\alpha} + \frac{1}{\beta} \)
(iii) \( (\alpha - \beta)^2 \)
(iv) \( \frac{1}{\alpha^2} + \frac{1}{\beta^2} \)
(v) \( \alpha^3 + \beta^3 \)
Answer: We have \( p(x) = 2x^2 - 4x + 5 \). Sum \( \alpha + \beta = 2 \), Product \( \alpha\beta = \frac{5}{2} \).
(i) \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 4 - 5 = -1 \).
(ii) \( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{2}{5/2} = \frac{4}{5} \).
(iii) \( (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = 4 - 10 = -6 \).
(iv) \( \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} = \frac{-1}{(5/2)^2} = -\frac{4}{25} \).
(v) \( \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = 8 - 15 = -7 \).

Question. On dividing the polynomial \( 4x^4 - 5x^3 - 39x^2 - 46x - 2 \) by the polynomial \( g(x) \), the quotient is \( x^2 - 3x - 5 \) and the remainder is \( -5x + 8 \). Find the polynomial \( g(x) \).
Answer: \( 4x^4 - 5x^3 - 39x^2 - 46x - 2 = g(x)(x^2 - 3x - 5) + (-5x + 8) \)
\( g(x) = \frac{4x^4 - 5x^3 - 39x^2 - 41x - 10}{x^2 - 3x - 5} \)
Hence, \( g(x) = 4x^2 + 7x + 2 \)

Question. If the squared difference of the zeroes of the quadratic polynomial \( f(x) = x^2 + px + 45 \) is equal to 144, find the value of \( p \).
Answer: Sum \( \alpha + \beta = -p \), Product \( \alpha\beta = 45 \).
Given \( (\alpha - \beta)^2 = 144 \)
\( (\alpha + \beta)^2 - 4\alpha\beta = 144 \)
\( (-p)^2 - 4(45) = 144 \)
\( p^2 - 180 = 144 \)
\( p^2 = 324 \Rightarrow p = \pm 18 \)