# Free Factorisation 02 Practice Test - 8th Grade

Which of the following is/are factor(s) of
49p236?

A. (7p6)
B. (7p+6)
C. (5p+6)
D. (6p+7)

#### SOLUTION

Solution : A and B

The given expression can be written as,
49p236=(7p)2(6)2
[Using the identity: a2b2=(a+b)(ab)]
=(7p+6)(7p6)

Hence, (7p+6) and (7p6)
are the factors of 49p236.

On dividing (24xy2)(z2) by 6yz2, we get

A.

4xz

B.

4xy

C.

4yz

D.

4xyz

#### SOLUTION

Solution : B

Comparing the coefficients of the different elements and dividing, we get:

246=4

x1=x

y2y=y

z2z2=1

Hence, we have 24xy2z2÷6yz2=4xy.

If 3 is a factor of 14pq, then 3 is also a factor of 28p2q.

A.

True

B.

False

#### SOLUTION

Solution : A

14pq itself is a factor of 28p2q. Thus, any factor of 14pq will also be a factor of 28p2q.

Find the value of the given expression
(x3+2x2+3x)÷2x

A.

(x2+2x+3)2

B.

12

C.

(x2+2x+3)

D.

1

#### SOLUTION

Solution : A

Divide all the terms of the expression (x3+2x2+3x) separately by 2x
x32x=x22;
2x22x=x;
3x2x=32.
Therefore, (x3+2x2+3x)2x=x22+x+32=(x2+2x+3)2

The value of  x(x+1)(x+2)(x+3)x(x+1) is __________.

A.

(x+2)

B.

(x + 2) (x + 3)

C.

(x+3)

D.

x(x+1)

#### SOLUTION

Solution : B

Given,

x(x+1)(x+2)(x+3)x(x+1)
On cancelling the common terms x and (x+1) , we get
(x+2)(x+3)

The factors of  ax + bx - ay - by are __________.

A.

(a + b) and (x + y)

B.

(a - b) and (x - y)

C.

(a + b) and (x - y)

D.

(a - b) and (x + y)

#### SOLUTION

Solution : C

We need to regroup ax  + bx - ay - by

We get x(a+b) - y(a+b)

Taking (a+b) common from both the terms we get (a+b)(x-y)

Thus, (a+b) and (x-y) is the factors of  ax  + bx - ay - by

Find the common factor between
2t,3t2and 4.

A.

3

B.

2

C.

1

D.

t

#### SOLUTION

Solution : C

2t, 3t2 and 4 can be factorized as,
2t=1×2×t
3t2=1×3×t×t
4=1×2×2
Hence, 1 is the only common factor among all three.

Factorise:4y212y+9

A.

(7y5)(7y5)

B.

(5y3)(5y3)

C.

(2y3)(2y3)

D.

(2y5)(2y5)

#### SOLUTION

Solution : C

The given expression can be rewritten as,
(2y)22(2y)(3)+32.
Comparing with the identity
(ab)2=a22ab+b2
we get,
(2y)22(2y)(3)+32
=(2y3)2
=(2y3)(2y3)

Which among the following is not a factor of  4y(y+3)?

A.

y2 + 3y

B.

4y2

C.

4(y + 3)

D.

4y

#### SOLUTION

Solution : B

4y(y+3) can be written as

4×y×(y+3)

=4y×(y+3)=4(y+3)×y
=4×y(y+3)
=4×(y2+3y)
=4y2+12y

So, 4y2 is a term in the expansion of 4y(y+3). It is not a factor.

4x2 is a factor of 4x(x+3)

A.

True

B.

False

#### SOLUTION

Solution : B

4x(x+3) = 4x2 + 3
So, 4x2 is a term of a given expression. 4x2 + 3 is not divisible by 4x2.
So, 4x2 is not a factor of 4x2+3.
The factors of  4x(x+3) are
4,x and (x+3)