This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Factorisation of Polynomials and Algebraic Identities”.

1. x-1 is a factor of 4x^{2}-9x-6.

a) True

b) False

View Answer

Explanation: According to

**factor theorem**, x-a is a factor of p(x) if p(a) = 0.

Therefore x-1 is a factor of 4x

^{2}-9x-6 is a factor if p(1)=0.

p(1) = 4(1)

^{2}-9(1)-6 = 4 – 9 – 6

= -11 ≠ 0

Therefore, we can say that x-1 is not a factor of 4x

^{2}-9x-6.

2. Find the value of k, if x-3 is a factor of 5x^{3}-2x^{2}+x+k.

a) 50

b) 60

c) -60

d) -120

View Answer

Explanation: According to

**factor theorem**, x-a is a factor of p(x) if p(a) = 0.

Here, it is given that x-3 is a factor of 5x

^{3}-2x

^{2}+x+k.

Therefore, p(3) must be equal to zero.

p(3) = 5(3)

^{3}-2(3)

^{2}+3+k = 0

Therefore, 5(27) – 2(9) + 3 + k = 0

135 – 18 + 3 + k=0

120 + k = 0

Therefore, k= -120

3. What do we get after factorising x^{2}+6x-27?

a) (x+9)(x-3)

b) (x-9)(x+3)

c) (x-9)(x-3)

d) (x+9)(x+3)

View Answer

Explanation: To factorise x

^{2}+6x-27, we have to find two numbers ‘a’ and ‘b’ such that a+b=6 and a*b=27.

For that we have to find factors of -27, which are ±1, ±3, ±9.

Now we have to arrange two numbers from these numbers such that a+b=6 and a*b=27.

By considering this, we get two numbers +9 and -3

9 + (-3) = 6 and 9*-3 = -27

Now after manipulating terms, we get x

^{2}+9x- 3x-27.

x

^{2}+9x- 3x-27 = x(x+9)-3(x+9)

= (x+9)(x-3).

4. What do we get after factoring 49x^{2}-28xy+.4y^{2}?

a) (7x+2y)^{2}

b) (49x-4y)^{2}

c) (7x-2y)^{2}

d) (7x-28y)^{2}

View Answer

Explanation: We know that a

^{2}-2ab+.b

^{2}= (a-b)

^{2}

49x

^{2}-28xy+.4y

^{2}can also be written as (7x)

^{2}-2(7)(2)xy+(2y)

^{2}

Here, a = 7x and b = 2y.

Therefore, 49x

^{2}-28xy+.4y

^{2}= (7x)

^{2}-2(7)(2)xy+(2y)

^{2}

= (7x-2y)

^{2}.

5. 26*34 = __________ (calculate without direct calculation).

a) 900

b) 884

c) 916

d) 844

View Answer

Explanation: 26*34 can also be written as (30-4)*(30+4)

We know that (a-b)*(a+b) = a

^{2}-b

^{2}

Similarly, 26*34 = (30-4)*(30+4)

= 30

^{2}– 4

^{2}

= 900 – 16

= 884.

6. 27*29 = __________ (calculate without direct calculation).

a) 783

b) 753

c) 763

d) 793

View Answer

Explanation: We know that (a+b)*(a+c) = a

^{2}+ (b+c)a + bc

27*29 can also be written as (25+2)*(25+4)

Now using above identity, 27*29 = (25+2)*(25+4)

= 25

^{2}+ (2+4)25 + (4)(2)

= 625 + 6(25) + 8

= 625 + 150 + 8

= 783.

7. 95^{3} = __________ (calculate without direct calculation).

a) 856395

b) 857625

c) 857375

d) 852395

View Answer

Explanation: We know that (x-y)

^{3}= x

^{3}– y

^{3}– 3 xy (x-y).

95

^{3}can also be written as (100-5)

^{3}

Therefore, 95

^{3}= (100-5)

^{3}= (100)

^{3}– (5)

^{3}– 3(100)(5)(100-5)

= 1000000 – 125 – 1500(95)

= 1000000 – 125 – 142500

= 857375.

8. What do we get after expanding (p+3q-2z)^{2}?

a) p^{2} + 9q^{2} + 4z^{2} + 6pq – 12qz + 4zp

b) p^{2} + 9q^{2} + 4z^{2} + 12pq + 12qz + 4zp

c) p^{2} + 9q^{2} + 4z^{2} – 12pq – 12qz – 4zp

d) p^{2} + 9q^{2} + 4z^{2} + 6pq – 12qz – 4zp

View Answer

Explanation: We know that (a+b+c)

^{2}= a

^{2}+ b

^{2}+ c

^{2}+ 2ab + 2bc + 2ca

(p+3q-2z)

^{2}can also be written as (p+3q+(-2z))

^{2}.

Here, a = p, b = 3q and c = -2z

Therefore, using that formula we get (p+3q+(-2z))

^{2}

= p

^{2}+ (3q)

^{2}+ (-2z)

^{2}+ 2(p)(3q) + 2(3q)(-2z) + 2(-2z)(p)

= p

^{2}+ 9q

^{2}+ 4z

^{2}+ 6pq – 12qz – 4zp.

9. What do we get after factorising x^{3}+ 8y^{3}+ z^{3} – 4xyz?

a) (x + 2y + z) (x^{2} + 4y^{2} + z^{2} – 2xy – 2yz – zx)

b) (x + 2y + z) (x^{2} + 4y^{2} + z^{2} – 2xy – 2yz – zx)

c) (x + 2y + z) (x^{2} + 4y^{2} + z^{2} – 2xy – 2yz – zx)

d) (x + 2y + z) (x^{2} + 4y^{2}+ z^{2} – 2xy – 2yz – zx)

View Answer

Explanation: We know that a

^{3}+ b

^{3}+ c

^{3}– 3abc = (a+b-c)(a

^{2}+ b

^{2}+ c

^{2}– ab – bc – ca)

In x

^{3}+ 8y

^{3}+ y

^{3}– 6xyz, a=x, b=2y and c=z

By using the above equation, we get x

^{3}+ 8y

^{3}+ z

^{3}– 4xyz = (x+2y+z) (x

^{2}+ (2y)

^{2}+ z

^{2}– x(2y) – (2y)(z) – zx)

= (x + 2y + z) (x

^{2}+ 4y

^{2}+ z

^{2}– 2xy – 2yz – zx).

**Sanfoundry Global Education & Learning Series – Mathematics – Class 9**.

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