CBSE Class 10 Mathematics Polynomials Worksheet Set 01

Question. The zeroes of the polynomial x² – 3x – m(m + 3) are:
(a) m, m + 3
(b) -m, m + 3
(c) m, - (m + 3)
(d) -m, - (m + 3)
Answer : B

Question. Given that one of the zeroes of the cubic polynomial ax³ + bx² + cx + d is zero, the product of the other two zeroes is:
(a) –(a/c)
(b) c/a
(c) 0
(d) -(b/a) 
Answer : B

Question. The quadratic polynomial, the sum of whose zeroes is -5 and their product is 6, is
(a) x² + 5x + 6
(b) x² – 5x + 6
(c) x² – 5x – 6
(d) – x² + 5x + 6
Answer : A

Question. If the zeroes of the quadratic polynomial x² + (a + 1)x + b are 2 and -3, then:
(a) a = -7, b = -1
(b) a = 5, b = -1
(c) a = 2, b = –6
(d) a = 0, b = –6
Answer : D

Question. If one of the zeroes of the cubic polynomial x³ + ax² + bx + c is - 1, then the product of the other two zeroes is:
(a) b – a + 1
(b) b – a - 1
(c) a – b + 1
(d) a – b - 1
Answer : A

Question. If one of the zeroes of the quadratic polynomial x² + 3x + k is 2, then the value of k is
(a) 10
(b) -10
(c) –7
(d) –2
Answer : B

Question. The number of polynomials having zeroes as -2 and 5 is:
(a) 1
(b) 2
(c) 3
(d) more than 3
Answer : D

Question. The zeroes of the quadratic polynomial x² + kx + k, where k ≠ 0 ,
(a) cannot both be positive
(b) cannot both be negative
(c) are always unequal
(d) are always equal
Answer : A

Question. If the zeroes of the quadratic polynomial ax² + bx + c, where c ≠ 0 are equal, then:
(a) c and a have opposite signs
(b) c and b have opposite signs
(c) c and a have the same sign
(d) c and b have the same sign
Answer : C

Question. If a, b are the zeros of the polynomial 5x² – 7x + 2, then the sum of their reciprocal is:
(a) 7/2
(b) 7/5
(c) 2/5
(d) 14/25
Answer : A

Question. The degree of the polynomial (x + 1) (x² – x + x⁴ - 1) is:
(a) 2
(b) 3
(c) 4
(d) 5
Answer : D

Question. The zeroes of the quadratic polynomial x² + 99x + 127 are:
(a) both positive
(b) both negative
(c) one positive and one negative
(d) both equal
Answer : B

Question. If one of the zeroes of a quadratic polynomial  of the form x² + ax + b is the negative of the other, then it:
(a) has no linear term and the constant term is negative.
(b) has no linear term and the constant term is positive.
(c) can have a linear term but the constant term is negative.
(d) can have a linear term but the constant term is positive.
Answer : A

Very Short Answer Questions

Each of the following questions are of 1 mark.

Question. For what value of \( k \), is \( -3 \) a zero of the polynomial \( x^2 + 11x + k \)?
Answer: Let \( p(x) = x^2 + 11x + k \) be the given polynomial, since \( -3 \) is the zero of the polynomial \( p(x) \).
\(\therefore\) \( p(-3) = 0 \)
\( \implies \) \( (-3)^2 + 11(-3) + k = 0 \)
\( \implies \) \( 9 - 33 + k = 0 \)
\( \implies \) \( -24 + k = 0 \)
\( \implies \) \( k = 24 \)

Question. Form a quadratic polynomial, the sum and product of whose zeros are \( (-3) \) and 2 respectively. 
Answer: Quadratic polynomial is given by
\( x^2 - (\text{Sum of zeros}) . x + \text{Product of zeros} \)
\( = x^2 - (-3)x + 2 = x^2 + 3x + 2 \)

Question. Find the quadratic polynomial whose zeros are \( -3 \) and 4.
Answer: Sum of zeros \( = -3 + 4 = 1 \),
Product of zeros \( = -3 \times 4 = -12 \)
\(\therefore\) Required polynomial \( = x^2 - x - 12 \)

Question. Can a quadratic polynomial \( x^2 + kx + k \) have equal zeros for some odd integer \( k > 1 \)?
Answer: No, for equal zeros, \( k^2 - 4k = 0 \) or \( k = 0, 4 \)
\( \implies \) \( k \) should be even.

Question. If one zero of the quadratic polynomial \( x^2 - 5x - 6 \) is 6 then find the other zero.
Answer: Let \( \alpha, 6 \) be the zeros of given polynomial.
Then \( \alpha + 6 = 5 \)
\( \implies \) \( \alpha = -1 \).

Question. Check whether \( (x + 1) \) is a factor of the polynomial \( p(x) = x^3 + 4x + 5 \).
Answer: \( p(x) = x^3 + 4x + 5 \)
Putting \( x = -1 \), we have
\( p(-1) = (-1)^3 + 4(-1) + 5 \)
\( = -1 - 4 + 5 = 0 \)
\( \implies \) \( (x + 1) \) is a factor of the polynomial \( p(x) = x^3 + 4x + 5 \).
Yes, \( (x + 1) \) is a factor of the polynomial \( p(x) = x^3 + 4x + 5 \).

Short Answer Questions-I

Each of the following questions are of 2 marks.

Question. A teacher asked 10 of his students to write a polynomial in one variable on a paper and then to hand over the paper. The following were the answers given by the students:
\( 2x + 3, 3x^2 + 7x + 2, 4x^3 + 3x^2 + 2, x^3 + \sqrt{3x} + 7, 7x + \sqrt{7}, 5x^3 - 7x + 2, 2x^2 + 3 - \frac{5}{x}, 5x - \frac{1}{2}, ax^3 + bx^2 + cx + d, x + \frac{1}{x} \)
Answer the following questions:
(i) How many of the above ten, are not polynomials?
(ii) How many of the above ten, are quadratic polynomials?
Answer: (i) \( 3 [(x^3 + \sqrt{3x} + 7, 2x^2 + 3 - \frac{5}{x}, x + \frac{1}{x})] \) are not polynomials.
(ii) \( 1 [(3x^2 + 7x + 2)] \) is a quadratic polynomial.

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 polynomial
\(\therefore\) \( p(-1) = k(-1)^2 - 4(-1) + k = 0 \)
\( \implies \) \( k + 4 + k = 0 \)
\( \implies \) \( 2k + 4 = 0 \)
\( \implies \) \( 2k = -4 \)
\( \implies \) \( k = -2 \)

Question. Find the zeros of the polynomial \( p(x) = 4x^2 - 12x + 9 \).
Answer: \( p(x) = 4x^2 - 12x + 9 = (2x - 3)^2 \)
For zeros, \( p(x) = 0 \)
\( \implies \) \( (2x - 3)(2x - 3) = 0 \)
\( \implies \) \( x = \frac{3}{2}, \frac{3}{2} \)

Question. If \( \alpha \) and \( \beta \) are zeros of \( p(x) = x^2 + x - 1 \), then find \( \frac{1}{\alpha} + \frac{1}{\beta} \).
Answer: Here, \( \alpha + \beta = -1, \alpha\beta = -1 \), so \( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha\beta} = \frac{-1}{-1} = 1 \)

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

Question. If one of the zeros of the quadratic polynomial \( f(x) = 4x^2 - 8kx - 9 \) is equal in magnitude but opposite in sign of the other, find the value of \( k \).
Answer: Let one root of the given polynomial be \( \alpha \).
Then the other root \( = -\alpha \)
\(\therefore\) Sum of the roots \( = (-\alpha) + \alpha = 0 \)
\( \implies \) \( \frac{-b}{a} = 0 \)
or \( \frac{8k}{4} = 0 \)
or \( k = 0 \)

Question. If \( \alpha \) and \( \beta \) are zeros of the polynomial \( f(x) = ax^2 + bx + c \), then find \( \frac{1}{\alpha^2} + \frac{1}{\beta^2} \).
Answer: Given : \( f(x) = ax^2 + bx + c \)
\( \alpha + \beta = -\frac{b}{a} \) and \( \alpha\beta = \frac{c}{a} \)
\( \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\beta^2 + \alpha^2}{(\alpha\beta)^2} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2} \)
\( = \left[ \frac{b^2}{a^2} - \frac{2c}{a} \right] \times \frac{a^2}{c^2} = \frac{b^2 - 2ac}{a^2} \times \frac{a^2}{c^2} = \frac{b^2 - 2ac}{c^2} \)

Question. If the product of two zeros of the polynomial \( p(x) = 2x^3 + 6x^2 - 4x + 9 \) is 3, then find its third zero.
Answer: Let \( \alpha, \beta, \gamma \) be the roots of the given polynomial and \( \alpha\beta = 3 \).
Then \( \alpha\beta\gamma = \frac{-d}{a} = -\frac{9}{2} \)
\( \implies \) \( 3 \times \gamma = -\frac{9}{2} \)
or \( \gamma = \frac{-3}{2} \)

Short Answer Questions-II

Each of the following questions are of 3 marks.

Question. Find the value of \( k \) such that the polynomial \( x^2 - (k + 6)x + 2(2k - 1) \) has sum of its zeros equal to half of their product. [
Answer: Sum of zeros \( = k + 6 \)
Product of zeros \( = 2(2k - 1) \)
Hence \( k + 6 = \frac{1}{2} \times 2(2k - 1) \)
\(\therefore\) \( k = 7 \) [CBSE Marking Scheme 2019 (30/1/1)]

Question. If one zero of the polynomial \( 3x^2 - 8x + 2k + 1 \) is seven times the other, find the value of \( k \).
Answer: Let \( \alpha \) and \( \beta \) be the zeros of the polynomial. Then as per question \( \beta = 7\alpha \).
Now sum of zeros \( = \alpha + \beta = \alpha + 7\alpha = -\left( -\frac{8}{3} \right) \)
\( \implies \) \( 8\alpha = \frac{8}{3} \)
or \( \alpha = \frac{1}{3} \)
and \( \alpha\beta = \alpha \times 7\alpha = \frac{2k + 1}{3} \)
\( \implies \) \( 7\alpha^2 = \frac{2k + 1}{3} \)
\( \implies \) \( 7 \left( \frac{1}{3} \right)^2 = \frac{2k + 1}{3} \) \( (\because \alpha = \frac{1}{3}) \)
\( \implies \) \( \frac{7}{9} = \frac{2k + 1}{3} \)
\( \implies \) \( \frac{7}{3} = 2k + 1 \)
\( \implies \) \( \frac{7}{3} - 1 = 2k \)
\( \implies \) \( k = \frac{2}{3} \)

Question. Quadratic polynomial \( 2x^2 - 3x + 1 \) has zeros as \( \alpha \) and \( \beta \). Now form a quadratic polynomial whose zeros are \( 3\alpha \) and \( 3\beta \). 
Answer: It is given that \( \alpha \) and \( \beta \) are zeros of the polynomial \( 2x^2 - 3x + 1 \).
\(\therefore\) \( \alpha + \beta = \frac{-(-3)}{2} = \frac{3}{2} \) and \( \alpha\beta = \frac{1}{2} \)
Now, new quadratic polynomial whose zeros are \( 3\alpha \) and \( 3\beta \) is given by
\( x^2 - (\text{sum of zeros})x + \text{product of zeros} \)
\( = x^2 - (3\alpha + 3\beta)x + 3\alpha \times 3\beta \)
\( = x^2 - 3(\alpha + \beta)x + 9\alpha\beta \)
\( = x^2 - 3 \times \frac{3}{2} x + 9 \times \frac{1}{2} \)
\( = x^2 - \frac{9}{2}x + \frac{9}{2} = \frac{1}{2}(2x^2 - 9x + 9) \)

Question. If \( \frac{2}{3} \) and \( -3 \) are the zeros of the polynomial \( ax^2 + 7x + b \), then find the values of \( a \) and \( b \). 
Answer: \( ax^2 + 7x + b \)
Sum of zeros \( = -\frac{b}{a} = \frac{-7}{a} \)
But sum of given zeros \( = \frac{2}{3} - 3 = -\frac{7}{3} \)
Now, \( -\frac{7}{a} = -\frac{7}{3} \)
\( \implies \) \( a = 3 \)
Product of zeros \( = \frac{c}{a} = \frac{b}{a} \)
But product of given zeros \( = \frac{2}{3} \times (-3) = -2 \). Therefore, \( \frac{b}{3} = -2 \)
\( \implies \) \( b = -6 \)

Question. Find the quadratic polynomial whose zeros are reciprocal of the zeros of the polynomial \( f(x) = ax^2 + bx + c, a \neq 0, c \neq 0 \). 
Answer: \( f(x) = ax^2 + bx + c \)
\( \alpha + \beta = -\frac{b}{a}, \alpha\beta = \frac{c}{a} \)
New sum of zeroes \( = \frac{1}{\alpha} + \frac{1}{\beta} = -\frac{b}{c} \)
New product of zeroes \( = \frac{1}{\alpha} \times \frac{1}{\beta} = \frac{a}{c} \)
\(\therefore\) Required quadratic polynomial \( = x^2 + \frac{b}{c}x + \frac{a}{c} \) or \( cx^2 + bx + a \)

Question. Find the zeros of the quadratic polynomial \( 7y^2 - \frac{11}{3}y - \frac{2}{3} \) and verify the relationship between zeros and the coefficients. 
Answer: Given: \( p(y) = 7y^2 - \frac{11}{3}y - \frac{2}{3} \)
\( = \frac{1}{3}(21y^2 - 11y - 2) = \frac{1}{3}(21y^2 - 14y + 3y - 2) \)
\( = \frac{1}{3}[7y(3y - 2) + 1(3y - 2)] = \frac{1}{3}(7y + 1)(3y - 2) \)
Equating \( p(y) = 0 \)
\( \implies \) \( \frac{1}{3}(7y + 1)(3y - 2) = 0 \)
\( \implies \) \( y = \frac{-1}{7} \) and \( y = \frac{2}{3} \)
Now, Sum of zeros \( = \frac{-b}{a} = \frac{-(-11/3)}{7} = \frac{11}{21} \) and \( \frac{2}{3} - \frac{1}{7} = \frac{14 - 3}{21} = \frac{11}{21} \)
and product of zeros \( = \frac{c}{a} = \frac{-2/3}{7} = -\frac{2}{21} \) and \( \frac{2}{3} \times -\frac{1}{7} = -\frac{2}{21} \) Hence verified.

Question. Find the quadratic polynomial sum and product of whose zeros are \( -1 \) and \( -20 \) respectively. Also find the zeros of the polynomial so obtained. 
Answer: Let \( \alpha \) and \( \beta \) be the zeros of the quadratic polynomial.
\(\therefore\) Sum of zeros, \( \alpha + \beta = -1 \)
and product of zeros, \( \alpha . \beta = -20 \)
So, quadratic polynomial is given by
\( x^2 - (\alpha + \beta)x + \alpha\beta \)
\( = x^2 - (-1)x - 20 \)
\( = x^2 + x - 20 \)
Now, for zeros of this polynominal
\( x^2 + x - 20 = 0 \)
\( \implies \) \( x^2 + 5x - 4x - 20 = 0 \)
\( \implies \) \( x(x + 5) - 4(x + 5) = 0 \)
\( \implies \) \( (x + 5)(x - 4) = 0 \)
\( \implies \) \( x = -5, 4 \)
Zeros are \( -5 \) and 4.

Question. If \( \alpha, \beta, \gamma \) be zeros of polynomial \( 6x^3 + 3x^2 - 5x + 1 \), then find the value of \( \alpha^{-1} + \beta^{-1} + \gamma^{-1} \). 
Answer: \(\because\) \( p(x) = 6x^3 + 3x^2 - 5x + 1 \) so \( a = 6, b = 3, c = -5, d = 1 \)
\(\therefore\) \( \alpha, \beta \) and \( \gamma \) are zeros of the polynomial \( p(x) \).
\( \alpha + \beta + \gamma = \frac{-b}{a} = \frac{-3}{6} = \frac{-1}{2} \)
\( \alpha\beta + \alpha\gamma + \beta\gamma = \frac{c}{a} = \frac{-5}{6} \)
and \( \alpha\beta\gamma = \frac{-d}{a} = \frac{-1}{6} \)
Now \( \alpha^{-1} + \beta^{-1} + \gamma^{-1} = \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} \)
\( = \frac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma} = \frac{-5/6}{-1/6} = 5 \)

Long Answer Questions

Each of the following questions are of 5 marks.

Question. Find the zeros of the polynomial \( f(x) = x^3 - 12x^2 + 39x - 28 \), if it is given that the zeros are in AP. 
Answer: Let \( \alpha, \beta, \gamma \) are the zeros of \( f(x) \). If \( \alpha, \beta, \gamma \) are in AP, then,
\( \beta - \alpha = \gamma - \beta \)
\( \implies \) \( 2\beta = \alpha + \gamma \) ---(i)
\( \alpha + \beta + \gamma = \frac{-b}{a} = \frac{-(-12)}{1} = 12 \)
\( \implies \) \( \alpha + \gamma = 12 - \beta \) ---(ii)
From (i) and (ii)
\( 2\beta = 12 - \beta \) or \( 3\beta = 12 \) or \( \beta = 4 \)
Putting the value of \( \beta \) in (i), we have
\( 8 = \alpha + \gamma \) ---(iii)
\( \alpha\beta\gamma = \frac{-d}{a} = \frac{-(-28)}{1} = 28 \)
\( (\alpha\gamma)4 = 28 \) or \( \alpha\gamma = 7 \) or \( \gamma = \frac{7}{\alpha} \) ---(iv)
Putting the value of \( \gamma = \frac{7}{\alpha} \) in (iii), we get
\( 8 = \alpha + \frac{7}{\alpha} \)
\( \implies \) \( 8\alpha = \alpha^2 + 7 \)
\( \implies \) \( \alpha^2 - 8\alpha + 7 = 0 \)
\( \implies \) \( \alpha^2 - 7\alpha - 1\alpha + 7 = 0 \)
\( \implies \) \( \alpha(\alpha - 7) - 1(\alpha - 7) = 0 \)
\( \implies \) \( (\alpha - 1)(\alpha - 7) = 0 \)
\( \implies \) \( \alpha = 1 \) or \( \alpha = 7 \)
Putting \( \alpha = 1 \) in (iv), we get
\( \gamma = \frac{7}{1} = 7 \)
or \( \alpha = 1, \gamma = 7 \) and \( \beta = 4 \)
\(\therefore\) zeros are 1, 4, 7.
Putting \( \alpha = 7 \) in (iv), we get
\( \gamma = \frac{7}{7} = 1 \)
or \( \alpha = 7, \gamma = 1 \) and \( \beta = 4 \)
\(\therefore\) zeros are 7, 4, 1.
Hence zeros are 1, 4, 7 or 7, 4, 1.

Please click on below link to download CBSE Class 10 Mathematics Polynomials Worksheet Set A