Class 10 Maths MCQ Questions of Polynomials with Answers was Prepared Based on Latest Pattern of exams. Students can solve MCQ Questions with Answers to realize their preparation level. Practice more and more these Multiple Questions will assist understudies with scoring better stamps in the board tests.

Check out Multiple Choice Questions of Polynomials are given underneath: -

                           Practice Class 10 Maths MCQ Question of Polynomials

1. If a polynomial p(y) is divided by y + 2, then which of the following can be the remainder:

(a)y + 1
(b)2y + 3
(c) 5
(d)y – 1

2. If the polynomials \(ax^3 + 4x^2 + 3x – 4\) and \(x^3 – 4x + a\), leave the same remainder when divided by \((x – 3)\), then value of a is :

(a) 2b
(b) – 1
(c) 1
(d) – 2b

3. If -2 is a zero of \(p(x) = (ax³ + bx² + x – 6)\) and \(p(x)\) leaves a remainder 4 when divided by \((x – 2)\), then the values of a and b are (respectively):

(a) a = 2,b = 2
(b) a = 0,b = – 2
(c) a = 0, b = 2
(d) a = 0, b = 0

4. If \(x^{101} + 1001\) is divided by \(x + 1\), then remainder is:

(a) 0
(b) 1
(c) 1490
(d) 1000

5. The degree of the polynomial \((x + 1)(x^2 – x – x^4 +1)\) is:

(a) 2
(b) 3
(c) 4
(d) 5

6. Find the remainder when \(x^4+x^3-2x^2+x+1\) is divided by \(x-1\)

(a) 1
(b) 5
(c) 2
(d) 3

7. If a and b are the zeroes of the polynomial \(x^2-11x +30\), Find the value of \(a^3 + b^3\)

(a) 134
(b) 412
(c) 256
(d) 341

8. \(S(x) = px^2+(p-2)x +2\). If 2 is the zero of this polynomial,what is the value of p

(a) -1
(b) 1/2
(c) -1/2
(d) +1

9. If the zeroes of the quadratic polynomial \(ax^2 + bx + c, c ≠ 0\) 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

10. What are the zeroes of the Polynomial \(p(x)=x^2+7x+10\)

(a) (2) , (-5)
(b) (-2) , (5)
(c) (2) , (5)
(d) (-2) , (-5)

11. The graph of the polynomial \(ax^2 + bx + c\) is an upward parabola if

(a) a > 0
(b) a < 0
(b) a = 0
(d) None

12. A polynomial of degree 3 is called

(a) a linear polynomial
(b) a quadratic polynomial
(c) a cubic polynomial
(d) a biquadratic polynomial

13. Dividend is equal to

(a) divisor × quotient + remainder
(b) divisior × quotient
(c) divisior × quotient – remainder
(d) divisor × quotient × remainder

14. The graph of the polynomial \(ax^2 + bx + c\) is a downward parabola if

(a) a > 0
(b) a < 0
(c) a = 0
(d) a = 1

15. If α, β are the zeros of the polynomial \(f(x) = x^2 + x + 1\), then \(\frac{1}{\alpha}+\frac{1}{\beta}=\)

(a) 1
(b) – 1
(c) 0
(d) None of these

16. If one zero of the polynomial \(f(x) = (k^2 + 4) x^2 + 13x + 4k\) is reciprocal of the other, then k =

(a) 2
(b) – 2
(c) 1
(d) – 1

17. If the sum of the zeros of the polynomial \(f(x) = 2x^3 – 3kx^2 + 4x – 5\) is 6, then the value of k is

(a) 2
(b) 4
(C) – 2
(d) – 4

18. sum of the squares of the zeroes of the polynomial \(p(x) = x^2 + 7x – k\) is 25, find k.

(a) 12
(b) 49
(c) – 24
(d) – 12

19. Let, α, s, v be the zeroes of \(x^3 + 4x^2 + x- 6\) such that the product of two of the zeroes is 6. Find the third zero.

(a) 6
(b) 2
(c) 4
(d) 1

20. Find a and b so that the polynomial \(6x^4 + 8x^3 – 5x^2 + ax + b\) is exactly divisible by \(2x^2 – 5\).

(a) a = 20, b = – 25
(b) a = 4, b = – 5
(c) a = 20, b = 5
(d) a = – 20, b = – 25

Answers & Explanations

1. Answer: (c) 5

Explanation: When p(y) is divided by y + 2, then the degree of remainder < deg of (y + 2)

2. Answer: (b) – 1

Explanation: \(p(x) = ax^3 + 4x^2 + 3x – 4\)
\(q(x) = x^3 – 4x + a\)
p(3) = q(3)
a = – 1

3. Answer: (c) a = 0, b = 2

Explanation: If – 2 is a zero
\(\Rightarrow\)p(- 2) = 0
\(\Rightarrow\) – 2a + b = 2
Also, p(2) = 4
2a + b = 2
\(\Rightarrow\)a = 0 and b = 2

4. Answer: (d) 1000

Explanation: \(p(x)\) is divided by \(x + 1\)
\(p(- 1) = (-1^{101}) + 1001 = 1000\)

5. Answer: (d) 5

Explanation: Since the highest degree variable in first bracket is \(x\) and in second bracket is \(x^4\) on multiplying \(x\) with \(x^4\).the highest power we obtain is 5.

6. Answer: (c) 2

Explanation: Let \(P(x) =x^4+x^3-2x^2+x+1\)
Remainder when divide by \(x-1\)
\(P(1) = 1+ 1-2+1+1=2\)

7. Answer: (d) 341

Explanation: \(a^3 + b^3= (a+b) (a^2+b^2-ab)=(a+b) \{(a+b)^2 -3ab\}\)
Now a+b=-(-11)/1=11
ab=30
So \(a^3+b^3=11( 121 -90)=341\)

8. Answer: (c) -1/2

Explanation: \(S(2)=4p+0+2=0 \Rightarrow p=-1/2\)

9. Answer: (c) c and a have the same sign

Explanation: Given that, the zeroes of the quadratic polynomial \(ax^2 + bx + c , c ≠ 0\) are equal
Value of discriminant (D) has to be zero
\(b^2– 4ac = 0\)
\(b^2 = 4ac\)
Since L.H.S. \((b^2)\) can never be negative
R.H.S. also can never be negative.
a and c must have same sign.

10. Answer: (d) (-2) , (-5)

Explanation: \(p(x)=x^2+7x+10\)
\(\Rightarrow p(x) = x^2 + 5x +2x +10 = x(x +5) + 2(x+5)=(x+2)(x+5)\)
So roots are (-2) , (-5)

11. Answer: (a) a > 0

12. Answer: (c) a cubic polynomial

13. Answer: (a) divisor × quotient + remainder

14. Answer: (b) a < 0

15. Answer: (b) – 1

Explanation: Given,
\(f(x) = x^2 + x + 1\)
α = a
β = b
c = c
α + β = – b/a
= – 1/1
= – 1 …….. Equ.(i)
αβ = c/a = 1 ….. Equ.(ii)
Dividing (i) from (ii)
We get,
– 1/1 = – 1
So,
1/α + 1/β = – 1

16. Answer: (a) 2

Explanation: Given;

f(x) = (k² + 4) x² + 13x + 4k,

One zero of the polynomial is reciprocal of the other,

Let a be the one zero,

∴ The other zero will be 1/a

As we know that,

Product of the zeros = c/a = 4k/k² + 4

∴ 4k/k² + 4 = 1

⇒ 4k = k² + 4

⇒ k ² + 4 – 4k = 0

⇒ (k – 2)² = 0

⇒ k = 2

So the value of k is 2

17. Answer: (b) 4

Explanation: Given,

f(x) = 2x³ – 3kx² + 4x – 5

Sum of the zeros of the polynomial = 6

Let x, y and z be the zeroes than,

x + y + z = 6……Equation (i)

So,

x + y + z = – b/a

= – ( – 3k)/2

From Eq. (i) we get,

3k/2 = 6

k/2 = 6/3

k = 2×2 = 4

18. Answer: (d) – 12

Explanation: \(p(x) = x^2 + 7x – k\)
let α, s be the zeroes
α + s = – 7
αs = – k
\(α^2 + s^2 = 25\)
\((α^2 + s) – 2αs = 25\)
49 + 2k = 25
k = -12

19. Answer: (d) 1

Explanation: α s v = 6,
αs = 6(given)
\(\Rightarrow v = 6/6=1\)

20. Answer: (d) a = – 20, b = – 25

Explanation: Divide the given polynomial by \(2×2 – 5\) get the remainder as \((20 + a)x + (b + 25)\) which should be zero.

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