# Free Factorisation 01 Practice Test - 8th Grade

Which of the following is the factorised form of  20l2m+30alm?

A. 10lm(l+3a)
B. 10lm(2l+3a)
C. 5lm(2l+3a)
D. 10lm(2l3a)

#### SOLUTION

Solution : B

Given, the expression is 20l2m+30alm.

The given expression can be factorised as follows :-

20l2m+30alm
=(2×2×5×l×l×m)+(2×3×5×a×l×m)

Taking out the common factors from both the terms, we get

=(2×5×l×m)[(2×l)+(3×a)]
=10lm(2l+3a)

Solve:4yz(z2+6z16)÷2y(z+8)

A.

z2

B.

z(z2)

C.

2z(z2)

D.

z(z4)

#### SOLUTION

Solution : C

Given, the expression is
4yz(z2+6z16)2y(z+8).

First, we will factorise (z2+6z16).

Comparing the above expression with the identity x2+(a+b)x+ab, we note that, (a+b)=6 and ab=16.

Since,
8+(2)=6 and (8)(2)=16,
the expression, z2+6z16
=z2+8z2z16
=z(z+8)2(z+8)
=(z2)(z+8) . . . (i)

By substituting (i) in the given expression, we get
=4yz(z2)(z+8)2y(z+8)
=4×y×z×(z2)×(z+8)2×y×(z+8)
=2z(z2)

Dividing x(3x227) by 3(x3) gives x2+3x.

A.

True

B.

False

#### SOLUTION

Solution : A

The given expression is

x(3x227)

Taking 3 common we get

= 3x(x29)

[Using a2b2=(a+b)(ab)]

= 3x(x+3)(x3)

= 3x(x+3)(x3)3(x3)

= x(x+3)

= x2+3x

Hence, the given statement is true

Dividing (a+b)2+(ab)2 by (a2+b2) gives

___

#### SOLUTION

Solution :

(a+b)2+(ab)2=(a2+b2+2ab)+(a2+b22ab)=2(a2+b2)

Dividing by a2+b2 we get,

= 2(a2+b2)(a2+b2)=2

Thus, ((a+b)2+(ab)2)(a2+b2)=2

Dividing 2x2+13x+15 by x+5 and then substituting x=2 in the resulting expression gives

___

#### SOLUTION

Solution :

2x2+13x+15=2x2+10x+3x+15=2x(x+5)+3(x+5)=(2x+3)(x+5)(x+5)=2x+3

Substituting x=2 in resulting expression gives 2×2+3=7.

Which of the following is/are the factor(s)
of 4x3+12x2+5x+15?

A.

(2x2+5)

B.

(4x2+5)

C.

(x+3)

D.

(2x2+3)

#### SOLUTION

Solution : B and C

Given, the expression is
4x3+12x2+5x+15.

Taking 4x2 common from the first two-terms and 5 common from the last two-terms, we get
4x2(x+3)+5(x+3)
=(x+3)(4x2+5)

Thus, (x+3) and (4x2+5)
are two factors of the given expression.

A factor of a22ab+b2c2 is ___________.

A.

ab

B.

abc

C.

a+b+c

D.

a+bc

#### SOLUTION

Solution : B

a22ab+b2c2=(ab)2c2
Using identity x2y2=(x+y)(xy)  we get

(ab)2c2 = [(ab)+c][(ab)c]

Hence, factors of a22ab+b2c2 are ab+c and abc

Divide (a2+7a+10) by (a+5).

A. 1
B. (a+1)
C. 2
D. (a+2)

#### SOLUTION

Solution : D

Given: a2+7a+10(a+5) ...(i)

Comparing a2+7a+10 with the identity x2+(a+b)x+ab,
we note that,(a+b)=7 and ab=10
So, 5+2=7 and (5)(2)=10

Hence,
a2+7a+10
=a2+5a+2a+10
=a(a+5)+2(a+5)
=(a+2)(a+5)

From (i), we get
a2+7a+10(a+5)=(a+2)(a+5)(a+5)
=(a+2)

Divide m2+7m60 by (m5).

A.

4(m - 6)

B.

2(m + 6)

C.

(m + 12)

D.

(m - 12)

#### SOLUTION

Solution : C

Given, the expression is
m2+7m60(m5).
Now, comparing the numerator with the identity x2+(a+b)x+ab
We note that,
(a+b)=7 and ab=60.
So,
12+(5)=7 and (12)(5)=60.
Hence,
m2+7m60=m2+12m5m60=m(m+12)5(m+12)=(m5)(m+12)
Therefore,
m2+7m60(m5)=(m5)(m+12)(m5)=(m+12)

Factorise: 25+70x+49x2

A.

(5x+7)(5x+2)

B.

(7x+5)(5x+3)

C.

(7x+5)(7x+5)

D.

(3x+7)2

#### SOLUTION

Solution : C

Given, the expression is 25+70x+49x2.

The given expression can be rewritten as,
49x2+70x+25

Using the identity: (a+b)2=a2+2ab+b2,we get

49x2+70x+25=(7x)2+2(5)(7x)+52
=(7x+5)2
=(7x+5)(7x+5)

25+70x+49x2=(7x+5)(7x+5)