CBSE Class 10 Polynomials Sure Shot Questions Set 02

Very Short Answer

Question. If the sum of zeroes of the quadratic polynomial \( 3x^2 - kx + 6 \) is 3, then find the value of \( k \).
Answer: Sol. Here \( a = 3, b = -k, c = 6 \)
Sum of the zeroes, \( (\alpha + \beta) = -\frac{b}{a} = 3 \) ...[Given
\( \Rightarrow \frac{-(-k)}{3} = 3 \)
\( \therefore k = 9 \)

Question. If \( \alpha \) and \( \beta \) are the zeroes of the polynomial \( ax^2 + bx + c \), find the value of \( \alpha^2 + \beta^2 \).
Answer: Sol. \( \alpha + \beta = -\frac{b}{a} \), \( \alpha\beta = \frac{c}{a} \)
\( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \)
\( \alpha^2 + \beta^2 = \left(-\frac{b}{a}\right)^2 - 2\left(\frac{c}{a}\right) \)
\( \alpha^2 + \beta^2 = \frac{b^2}{a^2} - \frac{2c}{a} \)
\( \therefore \alpha^2 + \beta^2 = \frac{b^2 - 2ac}{a^2} \)

Question. If the sum of the zeroes of the polynomial \( p(x) = (k^2 - 14)x^2 - 2x - 12 \) is 1, then find the value of \( k \).
Answer: Sol. \( p(x) = (k^2 - 14)x^2 - 2x - 12 \)
Here \( a = k^2 - 14, b = -2, c = -12 \)
Sum of the zeroes, \( (\alpha + \beta) = 1 \) ...[Given
\( \Rightarrow -\frac{b}{a} = 1 \Rightarrow \frac{-(-2)}{k^2 - 14} = 1 \)
\( \Rightarrow k^2 - 14 = 2 \Rightarrow k^2 = 16 \)
\( \therefore k = \pm 4 \)

Question. If \( \alpha \) and \( \beta \) are the zeroes of a polynomial such that \( \alpha + \beta = -6 \) and \( \alpha\beta = 5 \), then find the polynomial.
Answer: Sol. Quadratic polynomial is \( x^2 - Sx + P = 0 \)
\( \Rightarrow x^2 - (-6)x + 5 = 0 \Rightarrow x^2 + 6x + 5 = 0 \)

Question. A quadratic polynomial, whose zeroes are –4 and –5, is .... .
Answer: Sol. \( x^2 + 9x + 20 \) is the required polynomial.

Short Answer-I 

Question. Find the condition that zeroes of polynomial \( p(x) = ax^2 + bx + c \) are reciprocal of each other.
Answer: Sol. Let \( \alpha \) and \( \frac{1}{\alpha} \) be the zeroes of \( P(x) \).
\( P(x) = ax^2 + bx + c \) ...[Given
Product of zeroes = \( \frac{c}{a} \)
\( \alpha \times \frac{1}{\alpha} = \frac{c}{a} \Rightarrow 1 = \frac{c}{a} \)
\( \therefore a = c \) (Required condition)
\( \therefore \) Coefficient of \( x^2 \) = Constant term

Question. Form a quadratic polynomial whose zeroes are \( 3 + \sqrt{2} \) and \( 3 - \sqrt{2} \).
Answer: Sol. Sum of zeroes,
\( S = (3 + \sqrt{2}) + (3 - \sqrt{2}) = 6 \)
Product of zeroes,
\( P = (3 + \sqrt{2}) \times (3 - \sqrt{2}) \)
\( = (3)^2 - (\sqrt{2})^2 = 9 - 2 = 7 \)
\( \therefore \) Quadratic polynomial = \( x^2 - Sx + P \)
\( = x^2 - 6x + 7 \)

Question. Find a quadratic polynomial, the sum and product of whose zeroes are \( \sqrt{3} \) and \( \frac{1}{\sqrt{3}} \) respectively.
Answer: Sol. Sum of zeroes, \( (S) = \sqrt{3} \)
Product of zeroes, \( (P) = \frac{1}{\sqrt{3}} \)
Quadratic polynomial = \( x^2 - Sx + P \)
\( \Rightarrow x^2 - \sqrt{3}x + \frac{1}{\sqrt{3}} \Rightarrow \frac{\sqrt{3}x^2 - \sqrt{3}.\sqrt{3}x + 1}{\sqrt{3}} \)
\( \Rightarrow \frac{1}{\sqrt{3}}(\sqrt{3}x^2 - 3x + 1) \) or \( \sqrt{3}x^2 - 3x + 1 = 0 \)

Question. Find a quadratic polynomial, the sum and product of whose zeroes are 0 and \( -\sqrt{2} \) respectively.
Answer: Sol. Quadratic polynomial is
\( x^2 - (\text{Sum of zeroes})x + (\text{Product of zeroes}) \)
\( x^2 - (0)x + (-\sqrt{2}) = x^2 - \sqrt{2} \)

Question. Find the zeroes of the quadratic polynomial \( \sqrt{3}x^2 - 8x + 4\sqrt{3} \).
Answer: Sol. We have, \( \sqrt{3}x^2 - 8x + 4\sqrt{3} = 0 \)
\( = \sqrt{3}x^2 - 2x - 6x + 4\sqrt{3} \)
\( = x(\sqrt{3}x - 2) - 2\sqrt{3}(\sqrt{3}x - 2) \)
\( = (x - 2\sqrt{3})(\sqrt{3}x - 2) \)
Zeroes are:
\( x - 2\sqrt{3} = 0 \) or \( \sqrt{3}x - 2 = 0 \)
\( x = 2\sqrt{3} \) or \( x = \frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \)
\( x = \frac{2\sqrt{3}}{3} \)
\( \therefore \) Zeroes are \( 2\sqrt{3} \) and \( \frac{2\sqrt{3}}{3} \).

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: Sol. We have, \( 2x^2 - 5x - 3 = 0 \)
\( = 2x^2 - 6x + x - 3 \)
\( = 2x(x - 3) + 1(x - 3) \)
\( = (x - 3)(2x + 1) \)
Zeroes are:
\( x - 3 = 0 \Rightarrow x = 3 \) or \( 2x + 1 = 0 \Rightarrow x = -\frac{1}{2} \)
Since the zeroes of required polynomial is double of given polynomial.
\( \therefore \) Zeroes of the required polynomial are:
\( 3 \times 2 \) and \( \left(-\frac{1}{2} \times 2\right) \), i.e., \( 6, -1 \)
Sum of zeroes, \( S = 6 + (-1) = 5 \)
Product of zeroes, \( P = 6 \times (-1) = -6 \)
\( \therefore \) Quadratic polynomial is
\( x^2 - Sx + P \Rightarrow x^2 - 5x - 6 \) ...(i)
Comparing (i) with \( x^2 + px + q \)
\( \therefore p = -5, q = -6 \)

Question. Can \( (x - 2) \) be the remainder on division of a polynomial \( p(x) \) by \( (2x + 3) \)? Justify your answer.
Answer: Sol. In case of division of a polynomial by another polynomial the degree of remainder (polynomial) is always less than that of divisor.
\( \therefore (x - 2) \) can not be the remainder when \( p(x) \) is divided by \( (2x + 3) \) as the degree is same.

Question. Find a quadratic polynomial whose zeroes are \( \frac{3 + \sqrt{5}}{5} \) and \( \frac{3 - \sqrt{5}}{5} \).
Answer: Sol. Sum of zeroes \( (\alpha + \beta) \),
\( S = \frac{3 + \sqrt{5}}{5} + \frac{3 - \sqrt{5}}{5} \)
\( = \frac{3 + \sqrt{5} + 3 - \sqrt{5}}{5} = \frac{6}{5} \)
Product of zeroes \( (\alpha \times \beta) \),
\( P = \left(\frac{3 + \sqrt{5}}{5}\right) \left(\frac{3 - \sqrt{5}}{5}\right) = \frac{9 - 5}{25} = \frac{4}{25} \)
Quadratic polynomial is \( x^2 - Sx + P = 0 \)
\( = x^2 - \frac{6}{5}x + \frac{4}{25} = \frac{25x^2 - 30x + 4}{25} \)
\( = \frac{1}{25}(25x^2 - 30x + 4) \) or \( k(25x^2 - 30x + 4) \) ...where \( [k \in R] \)

Question. Find the quadratic polynomial whose zeroes are –2 and –5. Verify the relationship between zeroes and coefficients of the polynomial.
Answer: Sol. Sum of zeroes, \( S = (-2) + (-5) = -7 \)
Product of zeroes, \( P = (-2)(-5) = 10 \)
Quadratic polynomial is \( x^2 - Sx + P = 0 \)
\( x^2 - (-7)x + 10 \Rightarrow x^2 + 7x + 10 \)
Verification:
Here \( a = 1, b = 7, c = 10 \)
Sum of zeroes = \( (-2) + (-5) = -7 \)
\( = \frac{-7}{1} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} = -\frac{b}{a} \)
Product of zeroes = \( (-2)(-5) = 10 \)
\( = \frac{10}{1} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} = \frac{c}{a} \)
\( \therefore \) Relationship holds.

Question. Find the zeroes of the quadratic polynomial \( 3x^2 - 75 \) and verify the relationship between the zeroes and the coefficients.
Answer: Sol. We have, \( 3x^2 - 75 = 3(x^2 - 25) \)
\( = 3(x^2 - 5^2) = 3(x - 5)(x + 5) \)
Zeroes are: \( x - 5 = 0 \Rightarrow x = 5 \) or \( x + 5 = 0 \Rightarrow x = -5 \)
Verification:
Here \( a = 3, b = 0, c = -75 \)
Sum of the zeroes = \( 5 + (-5) = 0 \)
\( = -\frac{0}{3} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} = -\frac{b}{a} \)
Product of the zeroes = \( 5(-5) = -25 \)
\( = \frac{-75}{3} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} = \frac{c}{a} \)

Question. Find the zeroes of \( p(x) = 2x^2 - x - 6 \) and verify the relationship of zeroes with these coefficients.
Answer: Sol. \( p(x) = 2x^2 - x - 6 \)
\( = 2x^2 - 4x + 3x - 6 \)
\( = 2x(x - 2) + 3(x - 2) \)
\( = (x - 2)(2x + 3) \)
Zeroes are: \( x - 2 = 0 \Rightarrow x = 2 \) or \( 2x + 3 = 0 \Rightarrow x = -\frac{3}{2} \)
Verification:
Here \( a = 2, b = -1, c = -6 \)
Sum of zeroes = \( 2 + \left(-\frac{3}{2}\right) = \frac{4 - 3}{2} = \frac{1}{2} \)
\( = \frac{-(-1)}{2} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} = -\frac{b}{a} \)
Product of zeroes = \( 2 \times \left(-\frac{3}{2}\right) = -3 = \frac{-6}{2} \)
\( = \frac{\text{Constant term}}{\text{Coefficient of } x^2} = \frac{c}{a} \)
\( \therefore \) Relationship holds.

Question. What must be subtracted from the polynomial \( f(x) = x^4 + 2x^3 - 13x^2 - 12x + 21 \) so that the resulting polynomial is exactly divisible by \( x^2 - 4x + 3 \)?
Answer: Sol. By performing long division of \( x^4 + 2x^3 - 13x^2 - 12x + 21 \) by \( x^2 - 4x + 3 \):
Quotient is \( x^2 + 6x + 8 \) and Remainder is \( 2x - 3 \).
\( \therefore (2x - 3) \) should be subtracted from \( x^4 + 2x^3 - 13x^2 - 12x + 21 \).

Short Answer-II 

Question. Verify whether 2, 3 and 1/2 are the zeroes of the polynomial \( p(x) = 2x^3 - 11x^2 + 17x - 6 \).
Answer: Sol. \( p(x) = 2x^3 - 11x^2 + 17x - 6 \)
When \( x = 2 \), \( p(2) = 2(2)^3 - 11(2)^2 + 17(2) - 6 = 16 - 44 + 34 - 6 = 0 \)
When \( x = 3 \), \( p(3) = 2(3)^3 - 11(3)^2 + 17(3) - 6 = 54 - 99 + 51 - 6 = 0 \)
When \( x = \frac{1}{2} \), \( p\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 - 11\left(\frac{1}{2}\right)^2 + 17\left(\frac{1}{2}\right) - 6 \)
\( = \frac{2}{8} - \frac{11}{4} + \frac{17}{2} - 6 = \frac{1}{4} - \frac{11}{4} + \frac{17}{2} - 6 \)
\( = \frac{1 - 11 + 34 - 24}{4} = \frac{0}{4} = 0 \)
Yes, \( x = 2, 3 \) and \( 1/2 \) all are the zeroes of the given polynomial.

Question. Show that 1/2 and -3/2 are the zeroes of the polynomial \( 4x^2 + 4x - 3 \) and verify the relationship between zeroes and coefficients of polynomial.
Answer: Sol. Let \( P(x) = 4x^2 + 4x - 3 \)
When \( x = \frac{1}{2} \), \( P\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right)^2 + 4\left(\frac{1}{2}\right) - 3 = 1 + 2 - 3 = 0 \)
When \( x = -\frac{3}{2} \), \( P\left(-\frac{3}{2}\right) = 4\left(-\frac{3}{2}\right)^2 + 4\left(-\frac{3}{2}\right) - 3 = 4\left(\frac{9}{4}\right) + 2(-3) - 3 = 9 - 6 - 3 = 0 \)
Hence \( 1/2 \) and \( -3/2 \) are the zeroes.
Verification: \( a = 4, b = 4, c = -3 \)
Sum of zeroes = \( \frac{1}{2} + \left(-\frac{3}{2}\right) = -1 = -\frac{4}{4} = -\frac{b}{a} \)
Product of zeroes = \( \frac{1}{2} \times \left(-\frac{3}{2}\right) = -\frac{3}{4} = \frac{c}{a} \)
Relationship holds.

Question. Find a quadratic polynomial, the sum and product of whose zeroes are –8 and 12 respectively. Hence find the zeroes.
Answer: Sol. Let Sum \( S = -8 \), Product \( P = 12 \)
Quadratic polynomial is \( x^2 - Sx + P = x^2 - (-8)x + 12 = x^2 + 8x + 12 \)
For zeroes: \( x^2 + 6x + 2x + 12 = x(x + 6) + 2(x + 6) = (x + 2)(x + 6) \)
Zeroes are \( x = -2 \) or \( x = -6 \).

Question. Find a quadratic polynomial, the sum and product of whose zeroes are 0 and \( -\frac{3}{5} \) respectively. Hence find the zeroes.
Answer: Sol. Quadratic polynomial \( = x^2 - (0)x + \left(-\frac{3}{5}\right) = x^2 - \frac{3}{5} \)
For zeroes: \( x^2 - \left(\sqrt{\frac{3}{5}}\right)^2 = \left(x - \sqrt{\frac{3}{5}}\right) \left(x + \sqrt{\frac{3}{5}}\right) \)
Zeroes are \( \pm \sqrt{\frac{3}{5}} = \pm \frac{\sqrt{15}}{5} \).

Question. Find the zeroes of the quadratic polynomial \( 6x^2 - 3 - 7x \) and verify the relationship between the zeroes and the coefficients of the polynomial.
Answer: Sol. We have, \( 6x^2 - 7x - 3 = (2x - 3)(3x + 1) \)
Zeroes are: \( x = 3/2 \) or \( x = -1/3 \)
Verification: \( a = 6, b = -7, c = -3 \)
Sum of zeroes = \( \frac{3}{2} + \left(-\frac{1}{3}\right) = \frac{9-2}{6} = \frac{7}{6} = -\frac{b}{a} \)
Product of zeroes = \( \frac{3}{2} \times \left(-\frac{1}{3}\right) = -\frac{3}{6} = \frac{c}{a} \)
Relationship holds.

Question. Find the zeroes of the quadratic polynomial \( f(x) = x^2 - 3x - 28 \) and verify the relationship between the zeroes and the coefficients of the polynomial.
Answer: Sol. \( p(x) = x^2 - 3x - 28 = (x - 7)(x + 4) \)
Zeroes are: \( x = 7 \) or \( x = -4 \)
Verification: \( a = 1, b = -3, c = -28 \)
Sum of zeroes = \( 7 + (-4) = 3 = -\frac{-3}{1} = -\frac{b}{a} \)
Product of zeroes = \( 7(-4) = -28 = \frac{-28}{1} = \frac{c}{a} \)
Relationship holds.

Question. If \( \alpha \) and \( \beta \) are the zeroes of the polynomial \( 6y^2 - 7y + 2 \), find a quadratic polynomial whose zeroes are \( 1/\alpha \) and \( 1/\beta \).
Answer: Sol. Given: \( 6y^2 - 7y + 2 \). Here \( a = 6, b = -7, c = 2 \)
\( \alpha + \beta = 7/6 \), \( \alpha\beta = 2/6 = 1/3 \)
For required polynomial:
Sum \( S = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{7/6}{1/3} = \frac{7}{2} \)
Product \( P = \frac{1}{\alpha} \times \frac{1}{\beta} = \frac{1}{\alpha\beta} = \frac{1}{1/3} = 3 \)
Required polynomial: \( x^2 - \frac{7}{2}x + 3 = \frac{1}{2}(2x^2 - 7x + 6) \).

Question. Divide \( 3x^2 + 5x - 1 \) by \( x + 2 \) and verify the division algorithm.
Answer: Sol. Dividing \( 3x^2 + 5x - 1 \) by \( x + 2 \):
Quotient = \( 3x - 1 \), Remainder = \( 1 \)
Verification: Divisor \( \times \) Quotient + Remainder
\( = (x + 2)(3x - 1) + 1 = 3x^2 - x + 6x - 2 + 1 = 3x^2 + 5x - 1 = \text{Dividend} \)

Question. On dividing \( 3x^3 + 4x^2 + 5x - 13 \) by a polynomial \( g(x) \) the quotient and remainder were \( 3x + 10 \) and \( 16x - 43 \) respectively. Find the polynomial \( g(x) \).
Answer: Sol. \( P(x) = g(x) \cdot q(x) + r(x) \)
\( 3x^3 + 4x^2 + 5x - 13 = g(x)(3x + 10) + (16x - 43) \)
\( 3x^3 + 4x^2 - 11x + 30 = g(x)(3x + 10) \)
\( g(x) = \frac{3x^3 + 4x^2 - 11x + 30}{3x + 10} \)
Performing division: \( g(x) = x^2 - 2x + 3 \).

Question. Check whether polynomial \( x - 1 \) is a factor of the polynomial \( x^3 - 8x^2 + 19x - 12 \). Verify by division algorithm.
Answer: Sol. Let \( P(x) = x^3 - 8x^2 + 19x - 12 \)
Put \( x = 1 \): \( P(1) = 1 - 8 + 19 - 12 = 0 \)
Since remainder is 0, \( (x - 1) \) is a factor.
Verification: Dividing \( x^3 - 8x^2 + 19x - 12 \) by \( x - 1 \) gives quotient \( x^2 - 7x + 12 \) and remainder \( 0 \).

Long Answer 

Question. Divide \( 2x^2 + 4x^3 + 5x - 6 \) by \( 2x^2 + 1 + 3x \) and verify the division algorithm.
Answer: Sol. Dividing \( 4x^3 + 2x^2 + 5x - 6 \) by \( 2x^2 + 3x + 1 \):
Quotient = \( 2x - 2 \), Remainder = \( 9x - 4 \)
Verification: \( (2x^2 + 3x + 1)(2x - 2) + 9x - 4 \)
\( = 4x^3 - 4x^2 + 6x^2 - 6x + 2x - 2 + 9x - 4 \)
\( = 4x^3 + 2x^2 + 5x - 6 = \text{Dividend} \)

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: Sol. Since \( x - \sqrt{5} \) is a factor, \( P(\sqrt{5}) = 0 \).
By division, \( \frac{x^3 - 3\sqrt{5}x^2 - 5x + 15\sqrt{5}}{x^2 - 5} = x - 3\sqrt{5} \)
Factors are \( (x - \sqrt{5})(x + \sqrt{5})(x - 3\sqrt{5}) \).
All zeroes are \( \sqrt{5}, -\sqrt{5} \) and \( 3\sqrt{5} \).

Question. If a polynomial \( x^4 + 5x^3 + 4x^2 - 10x - 12 \) has two zeroes as –2 and –3, then find the other zeroes.
Answer: Sol. Zeroes are \( -2, -3 \), so \( (x+2)(x+3) = x^2 + 5x + 6 \) is a factor.
Dividing \( x^4 + 5x^3 + 4x^2 - 10x - 12 \) by \( x^2 + 5x + 6 \) gives quotient \( x^2 - 2 \).
Zeroes of \( x^2 - 2 \) are \( x = \pm \sqrt{2} \).
Other zeroes are \( \sqrt{2} \) and \( -\sqrt{2} \).

Question. Find all the zeroes of the polynomial \( 8x^4 + 8x^3 - 18x^2 - 20x - 5 \), if it is given that two of its zeroes are \( \sqrt{5/2} \) and \( -\sqrt{5/2} \).
Answer: Sol. Factor = \( (x - \sqrt{5/2})(x + \sqrt{5/2}) = x^2 - 5/2 \).
Dividing \( 8x^4 + 8x^3 - 18x^2 - 20x - 5 \) by \( x^2 - 5/2 \) gives \( 8x^2 + 8x + 2 = 2(4x^2 + 4x + 1) = 2(2x + 1)^2 \).
Zeroes of \( (2x + 1)^2 \) are \( -1/2, -1/2 \).
All zeroes are \( \sqrt{5/2}, -\sqrt{5/2}, -1/2, -1/2 \).

Question. If \( p(x) = x^3 - 2x^2 + kx + 5 \) is divided by \( (x - 2) \), the remainder is 11. Find \( k \). Hence find all the zeroes of \( x^3 + kx^2 + 3x + 1 \).
Answer: Sol. \( p(2) = 11 \Rightarrow (2)^3 - 2(2)^2 + k(2) + 5 = 11 \)
\( 8 - 8 + 2k + 5 = 11 \Rightarrow 2k = 6 \Rightarrow k = 3 \).
New polynomial \( q(x) = x^3 + 3x^2 + 3x + 1 = (x + 1)^3 \).
Zeroes of \( (x + 1)^3 \) are \( -1, -1, -1 \).

Question. If \( \alpha \) and \( \beta \) are zeroes of \( p(x) = kx^2 + 4x + 4 \), such that \( \alpha^2 + \beta^2 = 24 \), find \( k \).
Answer: Sol. \( \alpha + \beta = -4/k, \alpha\beta = 4/k \)
\( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 24 \)
\( (-4/k)^2 - 2(4/k) = 24 \Rightarrow \frac{16}{k^2} - \frac{8}{k} = 24 \)
\( 16 - 8k = 24k^2 \Rightarrow 3k^2 + k - 2 = 0 \)
\( (k + 1)(3k - 2) = 0 \Rightarrow k = -1 \) or \( k = 2/3 \).

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: Sol. \( \alpha + \beta = -5/2, \alpha\beta = k/2 \)
\( (\alpha + \beta)^2 - \alpha\beta = \frac{21}{4} \)
\( (-5/2)^2 - k/2 = \frac{21}{4} \Rightarrow \frac{25}{4} - \frac{k}{2} = \frac{21}{4} \)
\( -k/2 = -1 \Rightarrow k = 2 \).

Question. What must be subtracted from \( p(x) = 8x^4 + 14x^3 - 2x^2 + 8x - 12 \) so that \( 4x^2 + 3x - 2 \) is factor of \( p(x) \)?
Answer: Sol. Dividing \( 8x^4 + 14x^3 - 2x^2 + 8x - 12 \) by \( 4x^2 + 3x - 2 \):
Quotient = \( 2x^2 + 2x - 1 \), Remainder = \( 15x - 14 \).
\( \therefore \) Polynomial to be subtracted is \( (15x - 14) \).

Question. Find the values of \( a \) and \( b \) so that \( x^4 + x^3 + 8x^2 + ax - b \) is divisible by \( x^2 + 1 \).
Answer: Sol. Dividing \( x^4 + x^3 + 8x^2 + ax - b \) by \( x^2 + 1 \):
Remainder obtained is \( (a - 1)x - b - 7 \).
Since it is divisible, Remainder = 0.
\( a - 1 = 0 \Rightarrow a = 1 \) and \( -b - 7 = 0 \Rightarrow b = -7 \).
\( \therefore a = 1, b = -7 \).

Question. If a polynomial \( 3x^4 - 4x^3 - 16x^2 + 15x + 14 \) is divided by another polynomial \( x^2 - 4 \), the remainder comes out to be \( px + q \). Find the value of \( p \) and \( q \).
Answer: Sol. Dividing \( 3x^4 - 4x^3 - 16x^2 + 15x + 14 \) by \( x^2 - 4 \):
Remainder is \( -x - 2 \).
Given remainder = \( px + q \).
Comparing, \( p = -1, q = -2 \).

Question. If the polynomial \( (x^4 + 2x^3 + 8x^2 + 12x + 18) \) is divided by another polynomial \( (x^2 + 5) \), the remainder comes out to be \( (px + q) \), find the values of \( p \) and \( q \).
Answer: Sol. Dividing \( x^4 + 2x^3 + 8x^2 + 12x + 18 \) by \( x^2 + 5 \):
Remainder is \( 2x + 3 \).
Given remainder = \( px + q \).
Comparing, \( p = 2, q = 3 \).