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MCQ Questions for Class 10 Maths Polynomials with Answers

Free PDF Download of CBSE Class 10 Maths Chapter 2 Polynomials Multiple Choice Questions with Answers. MCQ Questions for Class 10 Maths with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 10 Maths Polynomials MCQs with Answers to know their preparation level.

Class 10 Maths MCQs Chapter 2 Polynomials

1. If one zero of the quadratic polynomial x² + 3x + k is 2, then the value of k is
(a) 10
(b) -10
(c) 5
(d) -5

Answer

Answer: b


2. Given that two of the zeroes of the cubic poly-nomial ax3 + bx² + cx + d are 0, the third zero is

Answer

Answer: a


3. If one of the zeroes of the quadratic polynomial (k – 1) x² + kx + 1 is – 3, then the value of k is

Answer

Answer: a


4. A quadratic polynomial, whose zeroes are -3 and 4, is
(a) x²- x + 12
(b) x² + x + 12
(c) \(\frac{x^{2}}{2}-\frac{x}{2}-6\)
(d) 2x² + 2x – 24

Answer

Answer: c


5. 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

Answer: d


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

Answer

Answer: d


7. Given that one of the zeroes of the cubic polynomial ax3 + bx² + cx + d is zero, the product of the other two zeroes is

Answer

Answer: b


8. 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

Answer: a


9. 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

Answer: b


10. The zeroes of the quadratic polynomial x² + kx + k, k? 0,
(a) cannot both be positive
(b) cannot both be negative
(c) are always unequal
(d) are always equal

Answer

Answer: a


11. If the zeroes of the quadratic polynomial ax² + bx + c, 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

Answer: c


12. 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

Answer: a


13. Which of the following is not the graph of quadratic polynomial?

Answer

Answer: d


14. The number of polynomials having zeroes as 4 and 7 is
(a) 2
(b) 3
(c) 4
(d) more than 4

Answer

Answer: d


15. A quadratic polynomial, whose zeores are -4 and -5, is
(a) x²-9x + 20
(b) x² + 9x + 20
(c) x²-9x- 20
(d) x² + 9x- 20

Answer

Answer: b


16. The zeroes of the quadratic polynomial x² + 1750x + 175000 are
(a) both negative
(b) one positive and one negative
(c) both positive
(d) both equal

Answer

Answer: a


17. The zeroes of the quadratic polynomial x² – 15x + 50 are
(a) both negative
(b) one positive and one negative
(c) both positive
(d) both equal

Answer

Answer: c


18. The zeroes of the quadratic polynomial 3x² – 48 are
(a) both negative
(b) one positive and one negative
(c) both positive
(d) both equal

Answer

Answer: b


19. The zeroes of the quadratic polynomial x² – 18x + 81 are
(a) both negative
(b) one positive and one negative
(c) both positive and unequal
(d) both equal and positive

Answer

Answer: d


20. The zeroes of the quadratic polynomial x² + px + p, p ≠ 0 are
(a) both equal
(b) both cannot be positive
(c) both unequal
(d) both cannot be negative

Answer

Answer: b


21. If one of the zeroes of the quadratic polynomial (p – l)x² + px + 1 is -3, then the value of p is

Answer

Answer: b


22. If the zeroes of the quadratic polynomial Ax² + Bx + C, C # 0 are equal, then
(a) A and B have the same sign
(b) A and C have the same sign
(c) B and C have the same sign
(d) A and C have opposite signs

Answer

Answer: b


23. If x3 + 1 is divided by x² + 5, then the possible degree of quotient is
(a) 0
(b) 1
(c) 2
(d) 3

Answer

Answer: b


24. If x3 + 11 is divided by x² – 3, then the possible degree of remainder is
(a) 0
(b) 1
(c) 2
(d) less than 2

Answer

Answer: d


25. If x4 + 3x² + 7 is divided by 3x + 5, then the possible degrees of quotient and remainder are:
(a) 3, 0
(b) 4, 1
(c) 3, 1
(d) 4, 0

Answer

Answer: a


26. If x5 + 2x4 + x + 6 is divided by g(x), and quotient is x² + 5x + 7, then the possible degree of g(x) is:
(a) 4
(b) 2
(c) 3
(d) 5

Answer

Answer: c


27. If x5 + 2x4 + x + 6 is divided by g(x) and quo-tient is x² + 5x + 7, then the possible degree of remainder is:
(a) less than 1
(b) less than 2
(c) less than 3
(d) less than 4

Answer

Answer: c


28. What is the number of zeroes that a linear poly-nomial has/have:
(a) 0
(b) 1
(c) 2
(d) 3

Answer

Answer: b


29. What is the number(s) of zeroes that a quadratic polynomial has/have:
(a) 0
(b) 1
(c) 2
(d) 3

Answer

Answer: c


30. What is the number(s) of zeores that a cubic polynomial has/have:
(a) 0
(b) 1
(c) 2
(d) 3

Answer

Answer: d


31. If one of the zeroes of the cubic polynomial x3 + px² + qx + r is -1, then the product of the other two zeroes is
(a) p + q + 1
(b) p-q- 1
(c) q – p + 1
(d) q – p – 1

Answer

Answer: c


32. If one zero of the quadratic polynomial x² + 3x + b is 2, then the value of b is
(a) 10
(b) -8
(c) 9
(d) -10

Answer

Answer: d


33. If 1 is one of the zeroes of the polynomial x² + x + k, then the value of k is:
(a) 2
(b) -2
(c) 4
(d) -4

Answer

Answer: b


34. If p(x) is a polynomial of at least degree one and p(k) = 0, then k is known as
(a) value of p(x)
(b) zero of p(x)
(c) constant term of p{x)
(d) none of these

Answer/Explanation

Answer: b
Explaination:
Zero of p(x)
Let p(x) = ax + b
Put x = k
p(k) = ak + b = 0
∴ k is zero of p(x).


35. If one of the zeroes of the quadratic polynomial (k -1)x² + kx + 1 the value of k is [NCERT Exemplar Problems]

Answer/Explanation

Answer: c
Explaination:
(k – 1)² + kx + 1
One zero is – 3, so it must satisfy the equation and make it zero.
∴ (k- 1) (- 3)² + k(-3) + 1 =0
⇒ 9k – 9 – Ik + 1 = 0
⇒ 6k – 8 = 0
⇒ k = \(\frac{8}{6}\) = \(\frac{4}{3}\)


36. If the zeroes of the quadratic polynomial x² + (a + 1) x + b are 2 and -3, then [NCERT Exemplar Problem, CBSE 2011]
(a) a = -7, b = -1
(b) a = 5, b = -1
(c) a = 2, b = -6
(d) a = 0, b = -6

Answer/Explanation

Answer: d
Explaination:
x² + (a + 1)x + b
∵ x = 2 is a zero
and x = – 3 is another zero
∴ (2)² + (a + 1)² + 6 = 0
and (- 3)² + (a + 1) (- 3) + b = 0
⇒ 4 + 2cr + 2 + & = 0
and 9 – 3a – 3 + b = 0
⇒ 2a + b = – 6 …(i)
and – 3a + b = – 6 …(ii)
Solving (i) and (ii), we get 5a = 0
⇒ a = 0 and b = – 6.


37. Which of the following is not the graph of a quadratic polynomial?[NCERT Exemplar Problems]

Answer/Explanation

Answer: a
Explaination:
∵ It crosses the x-axis in three points.


38. Zeroes of a polynomial can be determined graphically. No. of zeroes of a polynomial is equal to no. of points where the graph of polynomial
(a) intersects y-axis
(b) intersects x-axis
(c) intersects y-axis or intersects x-axis
(d) none of these

Answer/Explanation

Answer: b
Explaination:
(b) Intersects x-axis.


39. If graph of a polynomial does not intersects the x-axis but intersects y-axis in one point, then no. of zeroes of the polynomial is equal to
(a) 0
(b) 1
(c) 0 or 1
(d) none of these

Answer/Explanation

Answer: a
Explaination:
(a) Zero


40. A polynomial of degree n has
(a) only 1 zero
(b) at least n zeroes
(c) atmost n zeroes
(d) more than n zeroes

Answer/Explanation

Answer: c
Explaination:
Maximum number of zeroes of a
polynomial = degree of the polynomial.


41. If p(x) = axr + bx + c, then –\(\frac{b}{a}\) is equal to
(a) 0
(b) 1
(c) product of zeroes
(d) sum of zeroes

Answer/Explanation

Answer: d
Explaination:
(d) Sum of zeroes = –\(\frac{b}{a}\)


42. If p(x) = ax² + bx + c one zero is and a + b + c = 0, then one zero is
(a) \(\frac{-b}{a}\)
(b) \(\frac{c}{a}\)
(c) \(\frac{b}{c}\)
(d) none of these

Answer/Explanation

Answer: b
Explaination:
p(1) = 0; a(1)² + b(1) + c = 0
⇒ a + b + c = 0
∴ one zero (α) = 1
αβ = product of zeroes = \(\frac{c}{a}\)
⇒ 1.β = \(\frac{c}{a}\)
⇒ β = \(\frac{c}{a}\)
∴ zeroes are 1 and \(\frac{c}{a}\)


43. If p{x) = ax2 + bx + one of the zeroes is c and a + c = b, then

Answer/Explanation

Answer: c
Explaination:
p(-1) = 0; a(-1)² + b(-1) + c = 0
⇒ a – b + C = 0,
∴ One zero (a) = -1
αα = product of zeroes = \(\frac{c}{a}\)
⇒ (-1).β = \(\frac{c}{a}\)
⇒ β = \(\frac{-c}{a}\)


44. The number of polynomials having zeroes as -2 and 5 is [NCERT Exemplar Problems]
(a) 1
(b) 2
(c) 3
(d) more than 3

Answer/Explanation

Answer: d
Explaination:
∵ x² – 3x – 10, 2x² – 6x – 20,
\(\frac{1}{2}\)x² – \(\frac{3}{2}\)x – 5, 3x² – 9x – 30 etc.,
have zeroes – 2 and 5.


45. Given that one of the zeroes of the cubic polynomial ax3 + bx2 + cx + d is zero, the product of the other two zeroes is [NCERT Exemplar Problems]

Answer/Explanation

Answer: b
Explaination:
∵ αβ + βγ + γα = \(\frac{c}{a}\)
Let, α = 0
So, 0 + βγ + 0 = \(\frac{c}{a}\)
⇒ βγ= \(\frac{c}{a}\)


46. 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/Explanation

Answer: a
Explaination:
f(x) = x² + ax + b
Given: zeroes are α and – α
Sum of zeroes = α – α = 0
∴ f(x) = x² + b, which is not linear
Product of zeroes = α(-α) = – α² = \(\frac{b}{1}\)
⇒ -α² = b
It is possible when, b < 0.
Hence, it has no linear term and the constant term is negative.


47. If 4x² – 6x – m is divisible by x – 3, the value of m is exact divisor of
(a) 9
(b) 45
(c) 20
(d) 36

Answer/Explanation

Answer: a
Explaination:
Here p(3) = 0
⇒ 4(3)² – 6 × 3-m = 0
⇒ 36 – 18 – m = 0
⇒ m=18
∴ Value of m is exactly divisible by 9.


48. Which one of the following statements is correct
(a) if x6 + 1 is divided by x + 1, then the remainder is -2.
(b) if x6 + 1 is divided by x – 1, then the remainder is 2.
(c) if x6 + 1 is divided by x + 1, then the remainder is 1.
(d) if x6 + 1 is divided by x – 1, then the remainder is -1.

Answer/Explanation

Answer: b
Explaination:
p(x) = x6 + 1
when divided by x – 1, then remainder = p(1)
∴ p(1) = 16 + 1 = 2


49. Consider the following statements
(i) x – 2 is a factor of x3 – 3x² + 4x – 4.
(ii) x + 1 is a factor of 2x3 + 4x + 6.
(iii) x – 1 is a factor of x5 + x4 – x3 + x² -x + 1.
In these statements
(a) 1 and 2 are correct
(b) 1, 2 and 3 are correct
(c) 2 and 3 are correct
(d) 1 and 3 are correct

Answer/Explanation

Answer: a
Explaination:
x – 2 is a factor of x3 – 3x² + 4x – 4
∵ remainder is zero Similarly x + 1 is a factor of 2x3 + 4x + 6
but x – 1 is not a factor of x5 + x4 – x3 + x² – x + 1
∵ remainder is not zero
∴ Statements 1 and 2 are correct.


50. If f(x) = 5x – 10 is divided by x – √2, then the remainder will be
(a) non zero rational number
(b) an irrational number
(c) 0
(d) \(f\left(\frac{1}{\sqrt{2}}\right)\)

Answer/Explanation

Answer: b
Explaination:
Remainder = f(√2) = 5 × √2 – 10
= an irrational number


51. Zeroes of p(z) = z² – 27 are ______ and ______ .

Answer/Explanation

Answer:
Explaination:
For zeroes z² – 27 = 0
⇒ z² = 27
⇒ z = ±√27
⇒ z = ± 3√3


52. Verify that x = 3 is a zero of the polynomial. p(x) = 2x3 – 5x² – 4x + 3

Answer/Explanation

Answer:
Explaination:
Here p(x) = 2x3 – 5x² – 4x + 3
∴ p{3) = 2(3)3 – 5 x (3)² -4 x 3 + 3
= 54 – 45- 12 + 3 = 0
∵ P(3) = 0
∴ x = 3 is a zero of p(x)


53. The graph of y =f(x) is given below. How many zeroes are there of f(x)?

Answer/Explanation

Answer:
Explaination:
Graph of y = f(x) intersect x-axis in one point only.
Therefore number of zeroes of f(x) is one.


54. The graph of y = f(x) is given, how many zeroes are there of f(x)?

Answer/Explanation

Answer:
Explaination:
∵ Graph y =f(x) does not intersect x-axis.
∴ f(x) has no zeroes.


55. The graph of y = f(x) is given below, for some polynomial f(x). Find the number of zeroes of f(x).

Answer/Explanation

Answer:
Explaination:
∵ Graph of f(x) intersects x-axis at three different points.
∴ Number of zeroes of f(x) = 3.


56. The graph of x = p(y) is given below, for some polynomial p(y). Find the number of zeroes of p(y).

Answer/Explanation

Answer:
Explaination:
∵ Graph of p(y) intersects y-axis in four different points.
∴ Numbers of zeroes = 4


57.
Graph of the polynomial p(x) =px² + 4x – 4 is given as above. Find the value of p.

Answer/Explanation

Answer:
Explaination:
Graph of p(x) touches the x-axis at (2, 0)
∴ x = 2 is a zero of the p(x)
⇒ p(2) = 0
⇒ p(2)² + 4 × 2 – 4 = 0
4p + 4 = 0
⇒ p = -1


58. If the product of the zeroes of x2 – 3kx + 2k1 – 1 is 7, then values of k are _____ and _____ .

Answer/Explanation

Answer:
Explaination:
Product of zeroes = 7
⇒ 2k² – 1 = 7
⇒ 2k² = 8
⇒ k² = 4
⇒ k = ± 2


59. If zeroes ofp(x) = 2x² -Ix + k are reciprocal of each other, then value of k is _____ .

Answer/Explanation

Answer:
Explaination:
∵ Zeroes are reciprocal of each other
∴ Product of zeroes = 1
⇒ \(\frac{k}{2}\) = 1
⇒ k = 2


60. Find the product of the zeroes of – 2x² + kx + 6.

Answer/Explanation

Answer:
Explaination:
Here a = – 2, b = k, c = 6
Product of zeroes = \(\frac{c}{a}\)
i. e., α × β = \(\frac{6}{-2}\) = -3


61. Find the sum of the zeroes of the given quadratic polynomial -3x² + k.

Answer/Explanation

Answer:
Explaination:
Since polynomial is -3x² + 0x + k
∴ a = -3, b = 0, c = k
and sum of zeroes = \(\frac{-b}{a}\)
i. e., α + β = \(\frac{-b}{a}\)
⇒ α + β = \(\frac{0}{-3}\) = 0


62. If one zero of the polynomial x² -4x+ 1 is 2 + √3, write the other zero.

Answer/Explanation

Answer:
Explaination:
Let other zero be α ,


63. Write the polynomial, the product and sum of whose zeroes are –\(\frac{9}{2}\) and –\(\frac{3}{2}\) respectively.

Answer/Explanation

Answer:
Explaination:


64. The value of m, in order that x² – mx – 2 is the quotient where x3 + 3x² – 4 is divided by x + 2 is ____ .

Answer/Explanation

Answer:
Explaination:

Since quotient = x² – mx – 2
∴ x² – mx – 2 = x² + x – 2
On comparison of coefficient of x, we get m = – 1.


65. If one factor of x3 + 7kx² – 4kx + 12 is (x + 3), then the value of k is ______ .

Answer/Explanation

Answer:
Explaination:
p(x) = x3 + Ike – 4kx + 12
p(-3) = (-3)3 + 7k(-3)² – 4k (- 3) + 12
∵ x + 3 is a factor
∴ p(-3) = 0
⇒ -27 + 63k + 12k+ 12 = 0
⇒ 75k – 15 = 0
⇒ k = \(\frac{1}{2}\)


66. A polynomial of degree five is divided by a quadratic polynomial. If it leaves a remainder, then find the degree of remainder.

Answer/Explanation

Answer:
Explaination:
Degree of remainder is always less than divisor.
∴ Degree of remainder will be less than 2.
Hence, degree is 1 or 0.


67. Check whether 3x – 7 is a factor of polynomial 6x3 + x² – 26x – 25?

Answer/Explanation

Answer:
Explaination:


68. If x3 + x² – ax + b is divisible by x² – x, write the values of a and b.

Answer/Explanation

Answer:
Explaination:
∵ p(x) = x3+xl-ax + b is divisible by x² -x.
x (x – 1) is a factor of p(x).
⇒ x = 0 and x = 1 are zeroes of p(x).
⇒ p(0) = 0
⇒ (0)3 + (0)² – a × 0 + b = 0
⇒ b = 0 and p(1) = 0
⇒ 13 + (1)² – a × 1 + b = 0
⇒ 2 – a + 0 — 0
⇒ a = 2


We hope the given MCQ Questions for Class 10 Maths Polynomials with Answers will help you. If you have any query regarding CBSE Class 10 Maths Chapter 2 Polynomials Multiple Choice Questions with Answers, drop a comment below and we will get back to you at the earliest.

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