# Mathematics Questions and Answers – Factorisation of Polynomials and Algebraic Identities

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

1. x-1 is a factor of 4x2-9x-6.
a) True
b) False

Explanation: According to factor theorem, x-a is a factor of p(x) if p(a) = 0.
Therefore x-1 is a factor of 4x2-9x-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 4x2-9x-6.

2. Find the value of k, if x-3 is a factor of 5x3-2x2+x+k.
a) 50
b) 60
c) -60
d) -120

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 5x3-2x2+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 x2+6x-27?
a) (x+9)(x-3)
b) (x-9)(x+3)
c) (x-9)(x-3)
d) (x+9)(x+3)

Explanation: To factorise x2+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 x2+9x- 3x-27.
x2+9x- 3x-27 = x(x+9)-3(x+9)
= (x+9)(x-3).
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4. What do we get after factoring 49x2-28xy+.4y2?
a) (7x+2y)2
b) (49x-4y)2
c) (7x-2y)2
d) (7x-28y)2

Explanation: We know that a2-2ab+.b2 = (a-b)2
49x2-28xy+.4y2 can also be written as (7x)2-2(7)(2)xy+(2y)2
Here, a = 7x and b = 2y.
Therefore, 49x2-28xy+.4y2 = (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

Explanation: 26*34 can also be written as (30-4)*(30+4)
We know that (a-b)*(a+b) = a2 -b2
Similarly, 26*34 = (30-4)*(30+4)
= 302 – 42
= 900 – 16
= 884.

6. 27*29 = __________ (calculate without direct calculation).
a) 783
b) 753
c) 763
d) 793

Explanation: We know that (a+b)*(a+c) = a2 + (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)
= 252 + (2+4)25 + (4)(2)
= 625 + 6(25) + 8
= 625 + 150 + 8
= 783.

7. 953 = __________ (calculate without direct calculation).
a) 856395
b) 857625
c) 857375
d) 852395

Explanation: We know that (x-y)3 = x3– y3 – 3 xy (x-y).
953 can also be written as (100-5)3
Therefore, 953= (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) p2 + 9q2 + 4z2 + 6pq – 12qz + 4zp
b) p2 + 9q2 + 4z2 + 12pq + 12qz + 4zp
c) p2 + 9q2 + 4z2 – 12pq – 12qz – 4zp
d) p2 + 9q2 + 4z2 + 6pq – 12qz – 4zp

Explanation: We know that (a+b+c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
(p+3q-2z)2can 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
= p2 + (3q)2 + (-2z)2 + 2(p)(3q) + 2(3q)(-2z) + 2(-2z)(p)
= p2 + 9q2 + 4z2 + 6pq – 12qz – 4zp.

9. What do we get after factorising x3+ 8y3+ z3 – 4xyz?
a) (x + 2y + z) (x2 + 4y2 + z2 – 2xy – 2yz – zx)
b) (x + 2y + z) (x2 + 4y2 + z2 – 2xy – 2yz – zx)
c) (x + 2y + z) (x2 + 4y2 + z2 – 2xy – 2yz – zx)
d) (x + 2y + z) (x2 + 4y2+ z2 – 2xy – 2yz – zx)

Explanation: We know that a3 + b3 + c3 – 3abc = (a+b-c)(a2 + b2 + c2 – ab – bc – ca)
In x3 + 8y3 + y3 – 6xyz, a=x, b=2y and c=z
By using the above equation, we get x3 + 8y3 + z3 – 4xyz = (x+2y+z) (x2 + (2y)2 + z2 – x(2y) – (2y)(z) – zx)
= (x + 2y + z) (x2 + 4y2 + z2 – 2xy – 2yz – zx).

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