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Polynomial MCQ Questions for Class 9 Maths with Answers & Solutions

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1 Polynomial Multiple Choice Questions with Answers & Solutions
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Polynomial Multiple Choice Questions with Answers & Solutions

1. If y – 2 and 2y – 1 are the factors of ky2 + 5y + r then :
(A) k = r = 0
(B) k = r = – 1
(C) k = r = – 2
(D) k = r = – 3
Solution

Ans. (C)
Sol. Let f(y) = ky2 + 5y + r
so f(2) = 0
\Rightarrow 4k + 10 + r = 0 …(1)
and f(1/2) = 0
\Rightarrow \frac{{\text{k}}}{{\text{4}}} + \frac{5}{2} + {\text{r}} = 0 …(2)
From (1) and (2) k = r = – 2

2. If x = \frac{4}{3} is a zero of the polynomial f(x) = 6x3 – 11x2 + kx – 20 then k =
(A) 18
(B) – 18
(C) – 19
(D) 19
Solution

Ans. (D)
Sol. f(x) = 6x3 – 11x2 + kx – 20
f(4/3) = 0
\Rightarrow {\text{6}}{\left( {\frac{{\text{4}}}{{\text{3}}}} \right)^3} - 11{\left( {\frac{4}{3}} \right)^2} + {\text{k}}{\text{.}}\frac{{\text{4}}}{{\text{3}}} - 20 = 0
k = 19

3. If (x – 1) and (x + 3) are factors of x3 – ax2 – 13x + b then (a, b) =
(A) (3, – 15)
(B) (15, 3)
(C) (–3, 15)
(D) (3, 15)
Solution

Ans. (D)
Sol. Let f(x) = x3 – ax2 – 13x + b
f(1) = 0
\Rightarrow 1 – a – 13 + b = 0 …… (1)
f(–3) = 0
\Rightarrow – 27 – 9a + 39 + b = 0 …… (2)
on solving (1) & (2), (a, b) = (3, 15)

4.The remainder when x5 – x4 – 5x3 + 3x2 – x + 8 is divided by 2x – 1 is

(A) \frac{{32}}{{243}}

(B) \frac{{ - 32}}{{243}}

(C) \frac{{243}}{{32}}

(D) \frac{{ - 243}}{{32}} Solution

Ans. (C)

Sol. Let f(x) = x5 – x4 – 5x3 + 3x2 – x + 8

f(1/2) = {\left( {\frac{1}{2}} \right)^5} - {\left( {\frac{1}{2}} \right)^4} - 5{\left( {\frac{1}{2}} \right)^3} + 3{\left( {\frac{1}{2}} \right)^2} - \frac{1}{2} + 8

 f(1/2) = \frac{{243}}{{32}}

5. Degree of a zero polynomial is :
(A) 0
(B) 1
(C) not defined
(D) none of these
Solution

Ans. (C)
Sol. Degree of a zero polynomial is not defined.

6. (62)3 – (54)3 – (8)3 is divisible by :
(A) 9 & 13
(B) 17 & 23
(C) 13 & 23
(D) 9 and 31
Solution

Ans. (D)
Sol. (62)3 – (54)3 – (8)3 = 3 × 62 × (–54) × (–8)
so it is divisible by 9 & 31

7.Factorize {a^{2x}} - {b^{2x}}:

(A) ({a^x} - {b^x})({a^2} - {b^2})

(B) ({a^x} - {b^x})({a^2} + {b^2})

(C) ({a^x} - {b^x})({a^x} + {b^x})

(D) ({a^x} - {b^x})({a^x} - {b^x}) Solution

Ans. (C) 
Sol. {a^{2x}} - {b^{2x}} = (ax)2 – (bx)2 
= (ax – bx) (ax + bx) 

8.What must be added to \frac{7}{{3x}} to make it 4x:
(A) \frac{{12{x^2}-7}}{{3x}}

(B) 4x

(C) 12x3 – 7x2

(D) None of these

Solution
Ans. (A) 
Sol. \frac{7}{{3x}} + t = 4x
 t = 4x-\frac{7}{{3x}} = \frac{{12{x^2}-7x}}{{3x}}

9. If a – b = 3 and a3 – b3 = 117, then a + b =
(A) 5
(B) 7
(C) 9
(D) 11
Solution

Ans. (B)
Sol. a – b = 3 …… (1)
& a3 – b3 = 117 …… (2)
on cubing equation (1)
(a – b)3 = 27
\Rightarrow a3 – b3 – 3ab (a – b) = 27
\Rightarrow 117 – 3ab (3) = 27
\Rightarrow \therefore  ab = 10
we know
(a + b)2 = (a – b)2 + 4ab
= 32 + 4 × 40 = 49
∴ a + b = 7

10. The value of a for which (x – a) is a factor of the polynomial f(x) = x5 – a2x3 + 2x + a – 3 is
(A) 1
(B) 0
(C) 2
(D) –1
Solution

Ans. (A)
Sol. f(x) = x5 – a2x3 + 2x + a – 3
\because (x – a) is a factor of f(x).
∴ f(a) = 0
\Rightarrow a5 – a2.a3 + 2.a + a – 3 = 0
\Rightarrow a5 – a5 + 3a – 3 = 0
\therefore a = 1

11.Which one is a polynomial in the following :

(A) x + \frac{1}{x}

(B) \sqrt 7 {x^3} + \sqrt 5

(C) \sqrt {-4} {y^2} + 7\sqrt y + 3

(D) None of these

Solution
Ans. (B)
Sol. Exponent of variable should be integer. Here, option (B) is having exponent of variable in integers.

12. Which one is the zero of the polynomial p(x) = 3x2 – 5x – 12 :
(A) 3
(B) 4
(C) 7
(D) 8
Solution

Ans. (A)
Sol. 3x2 – 5x – 12 = 3x2 – 9x + 4x – 12
= 3x(x – 3) + 4(x – 3)
= (3x + 4) (x – 3)
Here, x = \frac{{ - 4}}{3} 
or   x = 3

13.The remainder when p(x) = 5x3 + 4x2 + 3x – 6 is divided by \left( {x - \frac{1}{2}} \right) is :

(A) \frac{{ - 23}}{8}

(B) \frac{{17}}{8}

(C) 0
(D) \frac{{ - 35}}{8}

Solution

Ans. (A)

Sol. p(x) = 5x3 + 4x2 + 3x – 6

p\left( {\frac{1}{2}} \right) = 5{\left( {\frac{1}{2}} \right)^3} + 4{\left( {\frac{1}{2}} \right)^2} + 3\left( {\frac{1}{2}} \right) - 6

= 5 \times \frac{1}{8} + \frac{4}{4} + 3 \times \frac{1}{2} - 6

= \frac{{5 + 8 + 12 - 48}}{8}

= \frac{{ - 23}}{8}

14. Factor of 4x2 + 4x – 3 is :
(A) 2x – 1
(B) 2x + 3
(C) x – 3
(D) Both A and B
Solution

Ans. (D)
Sol. 4x2 + 4x – 3 = 4x2 + 6x – 2x – 3
= 2x (2x + 3) –1 (2x + 3)
= (2x – 1) (2x + 3)

15. The H.C.F of 8x3 – 32x2 + 40x –16 and 4x3 – 24x2 + 36x – 16 is:
(A) 4(x–1)2
(B) 4(x+1)2
(C) (x–1)2
(D) (x+2)2
Solution

Ans. (A)
Sol. Factors of (8x3 – 32x2 + 40x –16) are 8(x – 1) (x – 1)(x – 2)
and factors of (4x3 – 24x2 + 36x – 16) are 4(x – 1)(x – 1)(x – 4)
∴ H.C.F of the above polynomials is
= 4(x – 1)(x – 1)

16.If 4x4 – 3x3 – 3x2 + x – 7 is divided by (1 – 2x) then remainder will be :

(A) \frac{{57}}{8}

(B) \frac{{ - 59}}{8}

(C) \frac{{55}}{8}

(D) \frac{{ - 55}}{8} Solution

Ans. (B)

Sol. p(x) = 4x4 – 3x3 – 3x2 + x – 7
p\left( {\frac{1}{2}} \right) = 4{\left( {\frac{1}{2}} \right)^4} - 3{\left( {\frac{1}{2}} \right)^3} - 3{\left( {\frac{1}{2}} \right)^2} + \left( {\frac{1}{2}} \right) - 7

= \frac{{4 \times 1}}{{16}} - \frac{{3 \times 1}}{8} - \frac{3}{4} + \frac{1}{2} - 7

= \frac{{2 - 3 - 6 + 4 - 56}}{8}

= \frac{{ - 59}}{8}

17. The factors of (x2 + 4y2 + 4y – 4xy – 2x – 8) are :
(A) (x – 2y – 4) (x – 2y + 2)
(B) (x – y + 2) (x – 4y – 4)
(C) (x + 2y – 4) (x + 2y + 2)
(D) None of these
Solution

Ans. (A)
Sol. x2 + 4y2 + 4y – 4xy – 2x – 8
= (x – 2y)2 – 2(x – 2y) – 8
= t2 – 2t – 8
= (t – 4) (t + 2)
= (x – 2y – 4) (x – 2y + 2)

18. Find the remainder when p(x) = x3 – 5x is divided by g(x) = –1 + x :
(A) 1
(B) –4
(C) 2
(D) –3
Solution

Ans. (B)
Sol. p(x) = x3 – 5x
p(1) = (1)3 – 5(1)
= 1 – 5
= – 4

19. Simplify : (6x2 – 5y)2
(A) 36x4 + 25y2 + 60x2y
(B) 36x4 + 25y2 – 60x2y
(C) 36x4 – 25y2
(D) None of these
Solution

Ans. (B)
Sol. (6x2 – 5y)2 = (6x2)2 – 2(6x2) (5y) + (5y)2
= 36x4 – 60x2y + 25y2

20. The expansion of (x + y – z)2 is
(A) x2 + y2 + z2 + 2xy + 2yz + 2zx
(B) x2 + y2 – z2 – 2xy + yz + 2zx
(C) x2 + y2 + z2 + 2xy – 2yz – 2zx
(D) x2 + y2 – z2 + 2xy – 2yz – 2zx
Solution

Ans. (C)
Sol. (x + y – z)2 = x2 + y2 + z2 + 2xy – 2yz – 2zx

21. Factorize 27 – 125 p3 – 135p + 225p2 :

(A) {(3-5p)^3}

(B) (3-5p)\,\,{(3 + \,5p)^2}

(C) {(3-5p)^2}\,\,(3 + \,5p)

(D) {(3 + \,5p)^3} Solution

Ans. (A)

Sol. 27 – 125 p3 – 135p + 225p2 = (3)3 – (5p)3 – 3(3)2(5p) + 3(3) (5p)2
= (3 – 5p)3

22. Factorize (x – 2y)2 – 5(x – 2y) + 6 :
(A) (x – 2y – 3)(x – 2y – 2)
(B) (x – 2y – 3)(x – 2y + 2)
(C) (x – 2y + 3)(x – 2y + 2)
(D) (x – 2y + 3)(x – 2y – 2)
Solution

Ans. (A)
Sol. Let x – 2y = p
So, p2 – 5p + 6 = p2 – 3p – 2p + 6
= p(p – 3) – 2(p – 3)
= (p – 3) (p – 2)
putting p = x – 2y
\therefore (x – 2y – 3)(x – 2y – 2)

23. If (x – a) is the GCD of x2 – x – 6 and x2 + 3x – 18 then a = :
(A) 3
(B) 6
(C) 9
(D) 12
Solution

Ans. (A)
Sol. p(x) = x2 – x – 6 = (x – 3) (x + 2)
q(x) = x2 + 3x – 18 = (x + 6) (x – 3)
Hence, common factor is x – 3.
i.e., a = 3

24. If 562 – 512 = 5P, then P is equal to :
(A) 106
(B) 107
(C) 105
(D) 104
Solution

Ans. (B)
Sol. 562 – 512 = 5P
\Rightarrow (56 – 51) (56 + 51) = 5P
5(107) = 5P
\therefore P = 107

25. Find the zero of the polynomial x3 – \frac{x}{2} – 7 :
(A) –3
(B) 3
(C) 2
(D) –1
Solution

Ans. (C)
Sol. {x^3} - \frac{x}{2} - 7 by hit and trial method, check with the options.
Put x = 2 
(2)3 – \frac{{(2)}}{2} - 7 
= 8 – 1 – 7 
= 0
 

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