**Class 9 Maths MCQs -Chapter 2 Polynomial with Solutions**

Class 9 Maths MCQs Polynomial with Solutions are created for the purpose of helping the class 10 students for the preparation of Term 1 CBSE Board exam 2021. The practice of these MCQs are required to every students because the pattern of the CBSE Board exam has changed and all the questions in the question paper are of MCQs types . All the questions of chapter 2 Polynomial are selected here as per the CBSE sample paper of maths published by CBSE.The solutions of these MCQs of the chapter Polynomial are created here as per the CBSE norms and explained here in easiest way so every students can understand the solutions.

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**Q1. The value of the polynomial P(x) = 5x³ – 3x² + 7x – 10,when x = -1 is:**

**(a) 7 (b) 17 (c) -18 (d) -25**

Ans. The given polynomial P(x) =5x³ – 3x² + 7x – 10

The value of the polynomial when x =-1

P(-1) = 5(-1)³ -3(-1)² +7(-1) -10 = -5 -3 -7 -10 =-25

**Q2.One of the zeroes of the polynomial P(x) = x² – 1 are :**

**(a) -1 (b) -2 (c) 0 (d) -3**

Ans. (a) -1

The given polynomial is P(x) = x² – 1

**P(x) = x² – 1 = 0**

(x +1)(x -1) = 0

x = -1 and x = 1

Q3.The product of ( x + 1/x)(x -1/x)(x² + 1/x²) is :

(a) (x – 1/x)³ (b) x + 1/x³ (c) x^{4}-1/ x^{4 }(d) x^{4}+ 1/ x^{4}

Ans. The given expression is

(x + 1/x)(x -1/x)(x² + 1/x²)

= (x² – 1/x²)(x²+1/x²) [i.e a² -b² = (a +b)(a – b)]

= ( x^{4}-1/ x^{4}) [ i.e a² – b² = (a + b)( a – b)]

Q4.If x/y + y/x = -1 (x,y ≠ 0), then the value of x³ – y³ is

(a) 1 (b) (x -y)(x² – y² +1) (c) 0 (d) -1/2

Ans.(c) 0

The given equation is

x/y + y/x = -1

(x² + y²)/xy = -1

x² + y² = -xy….(i)

Cubing both sides

The value of x³ – y³ = (x -y)(x² +y² + xy)

Putting the value from equation (i)

x³ – y³ = ( x – y)( -xy +xy) =0

Q5. The value of 625² – 575² is

(a) 50000 (b)30000 (c)60000 (d) 40000

Ans.

625² – 575² = (625 + 575)(625 – 575) = 1200 × 50 =60000

Q6. If x² + kx + 6 = ( x +1)( x – 3),then value of k is

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

Ans. The given equation is

x² + kx + 6 = (x +2)( x +3)

x² + kx + 6 = x² + (2+3)x + 2×3 = x² + 5x +6

On comparing LHS and RHS

k = 5

Q7. The coefficient of x³ in ( 2x² + 2x)(5x³ – 9x²) is:

(a) -15 (b) -18 (c) 27 (d) -28

Ans. (2x² + 2x)(5x³ -9x²) for getting the coefficient of x³,multiplying 2x and -9x² i.e 2x×-9x² = -18x³

Hence the coefficient of x³ is -18

Q8. Find the remainder when the polynomial 3x³ – 8x² + 17x + 4 is divided by x:

(a) 2 (b) -4 (c) 4 (d) 0

Ans.(c) 4

The given polynomial is 3x³ – 8x² + 17x + 4

Putting the value x = 0

3(0)³ – 8(0)² +17(0) + 4 = 4

Q9.If 2x + y + 3z = 0, then the value of 8x³ + y³ + 27z³ is :

(a) 18xyz (b) -18xyz (c) 36xyz (d) -36xyz

Ans.(a) 18xyz

The given equation is

2x + y + 3x = 0

Applying the identity

a³ + b³ +c³ – 3abc = (a + b +c)(a² + b² + c² – ab – bc -ac)

(2x)³ + y³ + (3z)³ – 3×2x×3z = ( 2x + y +3z)(4x² + y² + 9z² -2xy – 3yz – 6xz)

Putting the value of 2x + y +3z = 0

8x³ + y³ + 27z³ = 0

Q10. If area of the rectangle is 4x² + 4x – 3,then its possible diensions are :

(a) 2x +1, 2x +3 (b) 3x -1, 2x +1 (c) 2x -3,2x +5 (d) 2x +3,2x -1

Ans.

4x² + 4x – 3

=4x² + 6x – 2x – 3

=2x( 2x + 3) -1(2x + 3)

=(2x + 3)(2x – 1)

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