Factorisation(8th Grade > Mathematics ) Questions and answers for exam Preparation

8th Grade > Mathematics

FACTORISATION MCQs

Total Questions : 57 | Page 3 of 6 pages

Question 21. Dividing

x (3 x^{2} - 27)

3 (x - 3)

gives

x^{2} + 3 x

True
False
2z(z−2)
z(z−4)

Discuss Question

Answer: Option A. -> True
:
A
The given expression is

x (3 x^{2} - 27)

Taking 3 common we get
=

3 x (x^{2} - 9)

[Using

a^{2} - b^{2} = (a + b) (a - b)

]
=

3 x (x + 3) (x - 3)

\frac{3 x (x + 3) (x - 3)}{3 (x - 3)}

x (x + 3)

x^{2} + 3 x

Hence, the given statement is true

Question 22. A factor of

a^{2} - 2 a b + b^{2} - c^{2}

is ___________.

a−b
a−b−c
a+b+c
a+b−c

Discuss Question

Answer: Option B. -> a−b−c
:
B

a^{2} - 2 a b + b^{2} - c^{2} = (a - b)^{2} - c^{2}

Using identity

x^{2} - y^{2} = (x + y) (x - y)

we get

(a - b)^{2} - c^{2}

[(a - b) + c] [(a - b) - c]

Hence, factors of

a^{2} - 2 a b + b^{2} - c^{2}

are

a - b + c

and

a - b - c

Question 23.

Divide (a^{2} + 7 a + 10) by (a + 5) .

1
(a+1)
2
(a+2)

Discuss Question

Answer: Option D. -> (a+2)
:
D

Given: \frac{a^{2} + 7 a + 10}{(a + 5)} . . . (i)

Comparing

a^{2} + 7 a + 10

with the identity

x^{2} + (a + b) x + a b,

we note that, (a + b) = 7 a n d a b = 10

S o, 5 + 2 = 7 a n d (5) (2) = 10

Hence,

a^{2} + 7 a + 10

= a^{2} + 5 a + 2 a + 10

= a (a + 5) + 2 (a + 5)

= (a + 2) (a + 5)

From (i), we get

\frac{a^{2} + 7 a + 10}{(a + 5)} = \frac{(a + 2) (a + 5)}{(a + 5)}

= (a + 2)

Question 24.

4 x^{2}

is a factor of

4 x

(

x + 3

)

True
False
4(y + 3)
4y

Discuss Question

Answer: Option B. -> False
:
B

4 x

(

x + 3

) =

4 x^{2}

+ 3
So,

4 x^{2}

is a term of a given expression.

4 x^{2}

+ 3 is not divisible by

4 x^{2}

.
So,

4 x^{2}

is not a factor of

4 x^{2} + 3

.
The factors of

4 x

(

x + 3

) are

4, x a n d (x + 3)

Question 25.

F a c t o r i s e : 4 y^{2} - 12 y + 9

(7y−5)(7y−5)
(5y−3)(5y−3)
(2y−3)(2y−3)
(2y−5)(2y−5)

Discuss Question

Answer: Option C. -> (2y−3)(2y−3)
:
C
The given expression can be rewritten as,

(2 y)^{2} - 2 (2 y) (3) + 3^{2} .

Comparing with the identity

(a - b)^{2} = a^{2} - 2 a b + b^{2}

we get,

(2 y)^{2} - 2 (2 y) (3) + 3^{2}

= (2 y - 3)^{2}

= (2 y - 3) (2 y - 3)

Question 26. The value of

\frac{x (x + 1) (x + 2) (x + 3)}{x (x + 1)}

is __________.

(x+2)
(x + 2) (x + 3)
(x+3)
x(x+1)

Discuss Question

Answer: Option B. -> (x + 2) (x + 3)
:
B
Given,

\frac{x (x + 1) (x + 2) (x + 3)}{x (x + 1)}

On cancelling the common terms

x a n d (x + 1)

, we get

\Rightarrow (x + 2) (x + 3)

Question 27. If 3 is a factor of

14 p q

, then 3 is also a factor of

28 p^{2} q

True
False
4yz
4xyz

Discuss Question

Answer: Option A. -> True
:
A

14 p q

itself is a factor of

28 p^{2} q

. Thus, any factor of

14 p q

will also be a factor of

28 p^{2} q

Question 28.

Which of the following is the factorised form of $20 l^{2} m + 30 a l m ?$

$10 l m (l + 3 a)$
$10 l m (2 l + 3 a)$
$5 l m (2 l + 3 a)$
$10 l m (2 l - 3 a)$

Discuss Question

Answer: Option B. ->

10 l m (2 l + 3 a)

:
B

$Given, the expression is 20 l^{2} m + 30 a l m .$
The given expression can be factorised as follows :-
$20 l^{2} m + 30 a l m$
$= (2 \times 2 \times 5 \times l \times l \times m) + (2 \times 3 \times 5 \times a \times l \times m)$
Taking out the common factors from both the terms, we get
$= (2 \times 5 \times l \times m) [(2 \times l) + (3 \times a)]$
$= 10 l m (2 l + 3 a)$

Question 29.

$S o l v e : 4 y z (z^{2} + 6 z - 16) \div 2 y (z + 8)$

$z - 2$
$z (z - 2)$
$2 z (z - 2)$
$z (z - 4)$

Discuss Question

Answer: Option C. ->

2 z (z - 2)

:
C

Given, the expression is
$\frac{4 y z (z^{2} + 6 z - 16)}{2 y (z + 8)} .$
First, we will factorise $(z^{2} + 6 z - 16) .$
Comparing the above expression with the identity $x^{2} + (a + b) x + a b$ , we note that, $(a + b) = 6 a n d a b = - 16.$
Since,
$8 + (- 2) = 6 a n d (8) (- 2) = - 16$ ,
the expression, $z^{2} + 6 z - 16$
$= z^{2} + 8 z - 2 z - 16$
$= z (z + 8) - 2 (z + 8)$
$= (z - 2) (z + 8)$ . . . (i)
By substituting (i) in the given expression, we get
$= \frac{4 y z (z - 2) (z + 8)}{2 y (z + 8)}$
$= \frac{4 \times y \times z \times (z - 2) \times (z + 8)}{2 \times y \times (z + 8)}$
$= 2 z (z - 2)$

Question 30.

Dividing $x (3 x^{2} - 27)$ by $3 (x - 3)$ gives $x^{2} + 3 x$ .

True
False
$2 z (z - 2)$
$z (z - 4)$

Discuss Question

Answer: Option A. -> True
:
A

The given expression is
$x (3 x^{2} - 27)$
Taking 3 common we get
= $3 x (x^{2} - 9)$
[Using $a^{2} - b^{2} = (a + b) (a - b)$ ]
= $3 x (x + 3) (x - 3)$
= $\frac{3 x (x + 3) (x - 3)}{3 (x - 3)}$
= $x (x + 3)$
= $x^{2} + 3 x$
Hence, the given statement is true