CBSE Class 10 Polynomials Sure Shot Questions Set C

Question. What is the coefficient of \( x^2 \) in the polynomial \( P(x) = 3x^3 + 10(x - x^2) - 5x^2 - 2 \)?
Answer: -15

Question. Find the value of the polynomial \( P(x) \) when \( x = 3 \). \( P(x) = x^2 - 4x + 7 \)
Answer: 4

Question. What is the value of \( f\left(-\frac{3}{2}\right) \) where \( f(x) = 4x^2 + 3x + \frac{7}{2} \)?
Answer: 8

Question. If \( f(x) = x^2 - 5x - 14 \), find the value of \( f(7) \).
Answer: 0

Question. Find the zeros of the polynomial \( x^2 - 15x - 34 \)
Answer: (17, -2)

Question. What should be added to the polynomial \( x^2 - 5x + 4 \), so that 3 is the zero of the polynomial?
Answer: 2

Question. Which of the numbers 3, 2, -2, 1 are zeros of the polynomial \( x^2 - 4 \)?
Answer: (2, -2)

Question. What should be subtracted from the polynomial \( x^2 - 16x + 30 \) so that \( x = 15 \) is a zero of the polynomial?
Answer: 15

Question. Find the quotient when \( x^2 - 7x + 12 \) is divided by \( (x - 3) \).
Answer: \( (x - 4) \)

Question. Find the polynomial whose zeros are \( \sqrt{2} \) and \( -\sqrt{2} \).
Answer: \( x^2 - 2 \)

Question. Find the value of \( a \) in the polynomial \( x^2 + ax - 30 \) if 5 is the zero of the given polynomial.
Answer: a = 1

Question. What will be the remainder if \( f(x) = x^3 + 4x^2 - 3x + 1 \) is divided by \( (x - 2) \)?
Answer: 19

Question. If \( f(x) = 3x^2 - 7x + 8 \), find \( f(-2) \)
Answer: 34

Question. In the polynomial \( 3x^2 - 4x - 7 \), find the sum of two zeros of the polynomial.
Answer: \( \frac{4}{3} \)

Question. What is the product of zeros of the polynomial \( x^2 - 11x + 30 \)?
Answer: 30

Question. Which quadratic polynomial have its zeros as \( \frac{1}{4} \) and \( \frac{3}{4} \)?
Answer: \( \frac{1}{16}(16x^2 - 16x + 3) \)

Question. What is the value of 'k' in the polynomial \( P(x) = x^2 + 11x + k \) if -4 is a zero of the polynomial?
Answer: k = 28

Question. Find both the zeros of the polynomial \( 2x^2 - 3x - 14 \)
Answer: \( \left(\frac{7}{2}, -2\right) \)

Question. If \( (x + 2) \) is a factor of \( x^2 + ax + 2b \) and \( a + b = 4 \) then what is the value of a and b?
Answer: a = 3, b = 1

Question. What is the quotient if \( x^3 - 1 \) is divided by \( x^2 + x + 1 \)?
Answer: \( (x - 1) \)

Question. For what value of x both the polynomials \( 3x^2 + 8x + 4 \) and \( x^2 - x - 6 \) becomes zero?
Answer: x = -2

Question. How many zeros does the polynomial \( f(x) = (x - 1)(x + 1)(x - 2) \) have in all?
Answer: 3

Question. Find all the zeros of the polynomial \( f(x) = (x - 3)(x^2 - 9x + 20) \).
Answer: 3, 4, 5

Question. Find sum and product of the zeros of the polynomial \( 3x^2 - 14x + 11 \)
Answer: \( \frac{14}{3}, \frac{11}{3} \)

Question. What is the quotient and remainder if \( x^2 - 8x + 15 \) is divided by \( (x - 3) \)?
Answer: quotient \( (x - 5) \), R = 0

Question. How many maximum number of zeros a quadratic polynomial can have?
Answer: Two

Question. If one of the zero of the polynomial \( P(x) = x^2 - 13x + 40 \) is 8, which is the other zero?
Answer: 5

Question. Find the zeros of the polynomial \( x^2 - 25 = 0 \). Also give the sum of zeros.
Answer: (5, -5), 0

Question. Which quadratic polynomial have the sum and product of roots as -15 and 50?
Answer: \( x^2 + 15x + 50 \)

Question. Find the polynomial whose roots are \( 2\sqrt{3} \) and \( 3\sqrt{3} \).
Answer: \( x^2 - 5\sqrt{3}x + 18 \)

Question. What is the sum and product of 3 roots of the cubic polynomial \( x^3 - 7x + 6 \)?
Answer: (0, -6)

Question. What is the three roots of the polynomial \( (x + 4) (x^2 - 6x + 8) \).
Answer: (2, 4, -4)

Question. Two roots of the polynomial \( x^3 + x^2 - 9x - 9 \) are 3 and -3. What is the third root?
Answer: (-1)

Question. Find the quadratic polynomial whose two roots (zeros) are \( 3 + \sqrt{5} \) and \( 3 - \sqrt{5} \).
Answer: \( x^2 - 6x + 4 \)

Question. Complete the following :– Dividend = Divisor x __________ + __________
Answer: Quotient, Remainder

Question. \( (x - 1) \) is a factor of \( (x^3 + ax^2 + bx - 11) \) and \( a - b = 6 \). Find a and b.
Answer: a = 8, b = 2

Question. For what value of a, \( (x = 6) \) is a zero of the polynomial \( x^2 - ax - 6 \)?
Answer: a = 5

Question. Which factor is common in \( x^2 - 1 \), \( x^4 - 1 \) and \( (x - 1)^2 \)?
Answer: (x - 1)

Question. Find the common zero of \( (x^2 + 2x + 1) \), \( (x^2 - 1) \) and \( x^3 + 1 \).
Answer: -1

Question. Which is the factor common in \( 6(x + 1)(x + 2) \) and \( 9(x^3 + 1) \)?
Answer: \( 3(x + 1) \)

Question. For what value of x, both the polynomials \( x^2 - 3x + 2 \) and \( x^2 - 6x + 5 \) becomes zero?
Answer: x = 1

Question. For what value of k, \( (x = -4) \) is a zero of the polynomial \( 2x^2 + kx - 12 \)?
Answer: k = 5

Question. Which factor is common in \( x^2 + 8x + 15 \) and \( x^2 + 3x - 10 \)?
Answer: (x + 5)

Question. Reduce \( \frac{x^2 + 5x + 4}{x^2 + 2x + 1} \) to lowest terms.
Answer: \( \frac{x + 4}{x + 1} \)

Question. Find the cubic polynomial whose three zeros are 0, 4, -4.
Answer: \( x^3 - 16x \)

Question. If \( P(x) = x^2 + 5x + 2 \), what is the value of \( P(3) + P(2) \)?
Answer: 42

Short Answer Type Questions

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:

Question. \( 7y^2 - \frac{11}{3}y - \frac{2}{3} \)
Answer: Let \( f(y) = 7y^2 - \frac{11}{3}y - \frac{2}{3} \)
\( = \frac{1}{3}(21y^2 - 11y - 2) \)
\( = \frac{1}{3}(21y^2 - 14y + 3y - 2) \)
\( = \frac{1}{3}[7y(3y - 2) + 1(3y - 2)] \)
\( = \frac{1}{3}(3y - 2)(7y + 1) \)
Zeroes are:
\( 3y - 2 = 0 \Rightarrow y = \frac{2}{3} \) or \( 7y + 1 = 0 \Rightarrow y = -\frac{1}{7} \)
Verification:
Sum of zeroes = \( \frac{2}{3} + \left(-\frac{1}{7}\right) = \frac{2}{3} - \frac{1}{7} = \frac{14 - 3}{21} = \frac{11}{21} \)
\( = \frac{-b}{a} = \frac{-(\text{Coefficient of } y)}{(\text{Coefficient of } y^2)} = \frac{-(-11/3)}{7} = \frac{11}{21} \)
Product of zeroes = \( \left(\frac{2}{3}\right)\left(-\frac{1}{7}\right) = \frac{-2}{21} \)
\( = \frac{c}{a} = \frac{\text{Constant term}}{\text{Coefficient of } y^2} = \frac{-2/3}{7} = \frac{-2}{21} \)
\( \therefore \) Relationship holds.

Question. \( 4x^2 + 5\sqrt{2}x - 3 \)
Answer: Let \( f(x) = 4x^2 + 5\sqrt{2}x - 3 \)
\( = 4x^2 + 6\sqrt{2}x - \sqrt{2}x - 3 \)
...[By splitting the middle term]
\( = 2\sqrt{2}x(\sqrt{2}x + 3) - 1(\sqrt{2}x + 3) \)
\( = (\sqrt{2}x + 3)(2\sqrt{2}x - 1) \)
Zeroes are:
\( \sqrt{2}x + 3 = 0 \Rightarrow x = \frac{-3}{\sqrt{2}} \) or \( 2\sqrt{2}x - 1 = 0 \Rightarrow x = \frac{1}{2\sqrt{2}} \)
Verification:
Sum of zeroes = \( -\frac{3}{\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{-6 + 1}{2\sqrt{2}} = \frac{-5}{2\sqrt{2}} = \frac{-5\sqrt{2}}{4} \)
\( = \frac{-b}{a} = \frac{-(\text{Coefficient of } x)}{(\text{Coefficient of } x^2)} = \frac{-5\sqrt{2}}{4} \)
Product of zeroes = \( -\frac{3}{\sqrt{2}} \times \frac{1}{2\sqrt{2}} = -\frac{3}{4} \)
\( = \frac{c}{a} = \frac{\text{Constant term}}{\text{Coefficient of } x^2} = \frac{-3}{4} \)
\( \therefore \) Relationship holds.

Question. \( y^2 + \frac{3}{2}\sqrt{5}y - 5 \)
Answer: Let \( f(y) = y^2 + \frac{3}{2}\sqrt{5}y - 5 \)
\( = \frac{1}{2}(2y^2 + 3\sqrt{5}y - 10) \)
\( = \frac{1}{2}(2y^2 + 4\sqrt{5}y - \sqrt{5}y - 10) \)
\( = \frac{1}{2}[2y(y + 2\sqrt{5}) - \sqrt{5}(y + 2\sqrt{5})] \)
\( = \frac{1}{2}(y + 2\sqrt{5})(2y - \sqrt{5}) \)
Zeroes are:
\( y + 2\sqrt{5} = 0 \Rightarrow y = -2\sqrt{5} \) or \( 2y - \sqrt{5} = 0 \Rightarrow y = \frac{\sqrt{5}}{2} \)
Verification:
Sum of zeroes = \( -2\sqrt{5} + \frac{\sqrt{5}}{2} = \frac{-4\sqrt{5} + \sqrt{5}}{2} = \frac{-3\sqrt{5}}{2} \)
\( = \frac{-b}{a} = \frac{-(\text{Coefficient of } y)}{(\text{Coefficient of } y^2)} = \frac{-(3\sqrt{5}/2)}{1} = \frac{-3\sqrt{5}}{2} \)
Product of zeroes = \( -2\sqrt{5} \times \frac{\sqrt{5}}{2} = \frac{-10}{2} = -5 \)
\( = \frac{c}{a} = \frac{\text{Constant term}}{\text{Coefficient of } y^2} = \frac{-5}{1} = -5 \)
\( \therefore \) Relationship holds.

Question. \( v^2 + 4\sqrt{3}v - 15 \)
Answer: Let \( f(v) = v^2 + 4\sqrt{3}v - 15 \)
\( = v^2 + 5\sqrt{3}v - \sqrt{3}v - 15 \)
\( = v(v + 5\sqrt{3}) - \sqrt{3}(v + 5\sqrt{3}) \)
\( = (v + 5\sqrt{3})(v - \sqrt{3}) \)
Zeroes are:
\( v + 5\sqrt{3} = 0 \Rightarrow v = -5\sqrt{3} \) or \( v - \sqrt{3} = 0 \Rightarrow v = \sqrt{3} \)
Verification:
Sum of zeroes = \( -5\sqrt{3} + \sqrt{3} = -4\sqrt{3} \)
\( = \frac{-b}{a} = \frac{-(\text{Coefficient of } v)}{(\text{Coefficient of } v^2)} = \frac{-4\sqrt{3}}{1} = -4\sqrt{3} \)
Product of zeroes = \( (-5\sqrt{3})(\sqrt{3}) = -15 \)
\( = \frac{c}{a} = \frac{\text{Constant term}}{\text{Coefficient of } v^2} = \frac{-15}{1} = -15 \)
\( \therefore \) Relationship holds.

Question. \( 2s^2 - (1 + 2\sqrt{2})s + \sqrt{2} \)
Answer: Let \( f(s) = 2s^2 - (1 + 2\sqrt{2})s + \sqrt{2} \)
\( = 2s^2 - s - 2\sqrt{2}s + \sqrt{2} \)
\( = s(2s - 1) - \sqrt{2}(2s - 1) \)
\( = (2s - 1)(s - \sqrt{2}) \)
Zeroes are:
\( 2s - 1 = 0 \Rightarrow s = \frac{1}{2} \) or \( s - \sqrt{2} = 0 \Rightarrow s = \sqrt{2} \)
Verification:
Sum of zeroes = \( \frac{1}{2} + \sqrt{2} = \frac{1 + 2\sqrt{2}}{2} \)
\( = \frac{-b}{a} = \frac{-[-(1 + 2\sqrt{2})]}{2} = \frac{1 + 2\sqrt{2}}{2} \)
Product of zeroes = \( \frac{1}{2} \times \sqrt{2} = \frac{\sqrt{2}}{2} \)
\( = \frac{c}{a} = \frac{\text{Constant term}}{\text{Coefficient of } s^2} = \frac{\sqrt{2}}{2} \)
\( \therefore \) Relationship holds.

Long Answer Type Questions

For each of the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also, find the zeroes of these polynomials by factorisation.

Question. –8/3, 4/3
Answer: Given: Sum of zeroes \( (S) = -\frac{8}{3} \)
Product of zeroes \( (P) = \frac{4}{3} \)
Required quadratic polynomial,
\( f(x) = x^2 - Sx + P = x^2 - \left(-\frac{8}{3}\right)x + \frac{4}{3} = \frac{3x^2 + 8x + 4}{3} \)
Let \( f(x) = 3x^2 + 8x + 4 = 0 \)
\( 3x^2 + 6x + 2x + 4 = 0 \)
\( 3x(x + 2) + 2(x + 2) = 0 \)
\( (x + 2)(3x + 2) = 0 \)
Zeroes are \( x = -2 \) and \( x = -\frac{2}{3} \).

Question. 21/8, 5/16
Answer: Given: Sum of zeroes \( (S) = \frac{21}{8} \)
Product of zeroes \( (P) = \frac{5}{16} \)
Required quadratic polynomial,
\( f(x) = x^2 - Sx + P = x^2 - \frac{21}{8}x + \frac{5}{16} = \frac{16x^2 - 42x + 5}{16} \)
Let \( f(x) = 16x^2 - 42x + 5 = 0 \)
\( 16x^2 - 40x - 2x + 5 = 0 \)
\( 8x(2x - 5) - 1(2x - 5) = 0 \)
\( (2x - 5)(8x - 1) = 0 \)
Zeroes are \( x = \frac{5}{2} \) and \( x = \frac{1}{8} \).

Question. \( -2\sqrt{3}, -9 \)
Answer: Given: Sum of zeroes \( (S) = -2\sqrt{3} \)
Product of zeroes \( (P) = -9 \)
Required quadratic polynomial,
\( f(x) = x^2 - Sx + P = x^2 + 2\sqrt{3}x - 9 \)
Let \( f(x) = x^2 + 2\sqrt{3}x - 9 = 0 \)
\( x^2 + 3\sqrt{3}x - \sqrt{3}x - 9 = 0 \)
\( x(x + 3\sqrt{3}) - \sqrt{3}(x + 3\sqrt{3}) = 0 \)
\( (x + 3\sqrt{3})(x - \sqrt{3}) = 0 \)
Zeroes are \( x = -3\sqrt{3} \) and \( x = \sqrt{3} \).

Question. \( -\frac{3}{2\sqrt{5}}, -\frac{1}{2} \)
Answer: Given: Sum of zeroes \( (S) = -\frac{3}{2\sqrt{5}} \)
Product of zeroes \( (P) = -\frac{1}{2} \)
Required quadratic polynomial,
\( f(x) = x^2 - Sx + P = x^2 + \frac{3}{2\sqrt{5}}x - \frac{1}{2} = \frac{2\sqrt{5}x^2 + 3x - \sqrt{5}}{2\sqrt{5}} \)
Let \( f(x) = 2\sqrt{5}x^2 + 3x - \sqrt{5} = 0 \)
\( 2\sqrt{5}x^2 + 5x - 2x - \sqrt{5} = 0 \)
\( \sqrt{5}x(2x + \sqrt{5}) - 1(2x + \sqrt{5}) = 0 \)
\( (2x + \sqrt{5})(\sqrt{5}x - 1) = 0 \)
Zeroes are \( x = -\frac{\sqrt{5}}{2} \) and \( x = \frac{1}{\sqrt{5}} \).

Question. If the zeroes of the cubic polynomial \( x^3 - 6x^2 + 3x + 10 \) are of the form \( a, a + b \) and \( a + 2b \) for some real number \( a \) and \( b \), find the values of \( a \) and \( b \) as well as the zeroes of the given polynomial.
Answer: Let \( f(x) = x^3 - 6x^2 + 3x + 10 \)
Given \( a, (a + b) \) and \( (a + 2b) \) are the zeroes of \( f(x) \).
We know that Sum of zeroes = \( \frac{-\text{Coefficient of } x^2}{\text{Coefficient of } x^3} \)
\( (a) + (a + b) + (a + 2b) = \frac{-(-6)}{1} = 6 \)
\( 3a + 3b = 6 \Rightarrow a + b = 2 \Rightarrow b = 2 - a \) ...(i)
Product of zeroes = \( \frac{-\text{Constant term}}{\text{Coefficient of } x^3} \)
\( (a)(a + b)(a + 2b) = \frac{-10}{1} = -10 \)
\( (a)(2)(2 + b) = -10 \) ...[Using (i)]
\( 2a(2 + 2 - a) = -10 \Rightarrow 2a(4 - a) = -10 \)
\( 4a - a^2 = -5 \Rightarrow a^2 - 4a - 5 = 0 \)
\( (a - 5)(a + 1) = 0 \Rightarrow a = 5 \) or \( a = -1 \)
If \( a = 5 \), \( b = 2 - 5 = -3 \)
If \( a = -1 \), \( b = 2 - (-1) = 3 \)
When \( a = -1 \) and \( b = 3 \), zeroes are \( -1, 2, 5 \).
When \( a = 5 \) and \( b = -3 \), zeroes are \( 5, 2, -1 \).
The values of \( (a, b) \) are \( (-1, 3) \) or \( (5, -3) \) and zeroes are \( -1, 2, 5 \).

Question. If \( \sqrt{2} \) is a zero of the cubic polynomial \( 6x^3 + \sqrt{2}x^2 - 10x - 4\sqrt{2} \), find its other two zeroes.
Answer: Let \( f(x) = 6x^3 + \sqrt{2}x^2 - 10x - 4\sqrt{2} \)
Since \( \sqrt{2} \) is a zero, \( (x - \sqrt{2}) \) is a factor.
By division algorithm, \( \frac{6x^3 + \sqrt{2}x^2 - 10x - 4\sqrt{2}}{x - \sqrt{2}} = 6x^2 + 7\sqrt{2}x + 4 \)
To find other zeroes, \( 6x^2 + 7\sqrt{2}x + 4 = 0 \)
\( 6x^2 + 4\sqrt{2}x + 3\sqrt{2}x + 4 = 0 \)
\( 2\sqrt{2}x(\frac{3}{\sqrt{2}}x + 2) + \sqrt{2}(\frac{3}{\sqrt{2}}x + 2) = 0 \)? No, let's re-factorize:
\( \sqrt{2}x(3\sqrt{2}x + 4) + 1(3\sqrt{2}x + 4) = 0 \)
\( (\sqrt{2}x + 1)(3\sqrt{2}x + 4) = 0 \)
Zeroes are \( x = -\frac{1}{\sqrt{2}} \) and \( x = -\frac{4}{3\sqrt{2}} \).

Question. Find \( k \) so that \( x^2 + 2x + k \) is a factor of \( 2x^4 + x^3 - 14x^2 + 5x + 6 \). Also, find all the zeroes of the two polynomials.
Answer: Dividing \( 2x^4 + x^3 - 14x^2 + 5x + 6 \) by \( x^2 + 2x + k \):
Remainder is \( (7k + 21)x + (2k^2 + 8k + 6) \).
For \( x^2 + 2x + k \) to be a factor, Remainder must be 0.
\( 7k + 21 = 0 \Rightarrow k = -3 \).
Also, \( 2(-3)^2 + 8(-3) + 6 = 18 - 24 + 6 = 0 \). Thus \( k = -3 \).
For zeroes of \( x^2 + 2x - 3 = (x + 3)(x - 1) \), zeroes are \( -3, 1 \).
For zeroes of \( 2x^4 + x^3 - 14x^2 + 5x + 6 \):
\( = (x^2 + 2x - 3)(2x^2 - 3x - 2) = (x+3)(x-1)(2x+1)(x-2) \).
Zeroes are \( -3, 1, -1/2, 2 \).

Question. If \( x - \sqrt{5} \) is a factor of the cubic polynomial \( x^3 - 3\sqrt{5}x^2 + 13x - 3\sqrt{5} \), then find all the zeroes of the polynomial.
Answer: Dividing \( x^3 - 3\sqrt{5}x^2 + 13x - 3\sqrt{5} \) by \( x - \sqrt{5} \) gives quotient \( x^2 - 2\sqrt{5}x + 3 \).
To find other zeroes, \( x^2 - 2\sqrt{5}x + 3 = 0 \).
Using splitting the middle term: \( x^2 - (\sqrt{5} + \sqrt{2} + \sqrt{5} - \sqrt{2})x + (\sqrt{5} + \sqrt{2})(\sqrt{5} - \sqrt{2}) = 0 \)
\( \{x - (\sqrt{5} + \sqrt{2})\}\{x - (\sqrt{5} - \sqrt{2})\} = 0 \)
Zeroes are \( \sqrt{5} \), \( \sqrt{5} + \sqrt{2} \) and \( \sqrt{5} - \sqrt{2} \).

Question. For which values of \( a \) and \( b \), the zeroes of \( q(x) = x^3 + 2x^2 + a \) are also the zeroes of the polynomial \( p(x) = x^5 - x^4 - 4x^3 + 3x^2 + 3x + b \)? Which zeroes of \( p(x) \) are not the zeroes of \( q(x) \)?
Answer: Dividing \( p(x) \) by \( q(x) \):
Quotient = \( x^2 - 3x + 2 \), Remainder = \( -(1+a)x^2 + (3+3a)x + (b-2a) \).
Remainder must be zero.
\( 1 + a = 0 \Rightarrow a = -1 \)
\( b - 2a = 0 \Rightarrow b = 2(-1) = -2 \)
For \( a = -1, b = -2 \), zeroes of \( q(x) \) are also zeroes of \( p(x) \).
Zeroes of \( p(x) \) which are not zeroes of \( q(x) \) are zeroes of the quotient \( x^2 - 3x + 2 \).
\( x^2 - 3x + 2 = (x - 2)(x - 1) = 0 \)
Hence, \( x = 2 \) and \( 1 \) are the zeroes not in \( q(x) \).