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CBSE Questions for Class 8 Maths Factorisation Quiz 5 - MCQExams.com
CBSE
Class 8 Maths
Factorisation
Quiz 5
Factorisation of the expression
x
4
y
z
+
x
y
4
z
+
x
y
z
4
results in:
Report Question
0%
x
y
z
(
x
2
+
y
2
+
z
2
)
0%
x
y
z
(
x
3
+
y
3
+
z
3
)
0%
x
3
y
3
z
3
(
x
+
y
+
z
)
0%
x
2
y
2
z
2
(
x
2
+
y
2
+
z
2
)
Explanation
x
4
y
z
+
x
y
4
z
+
x
y
z
4
=
x
y
z
(
x
3
+
y
3
+
z
3
)
Factorise :
12
x
2
−
24
y
2
+
36
z
2
Report Question
0%
12
x
2
−
2
y
2
+
3
z
2
0%
12
(
x
2
−
2
y
2
+
3
z
2
)
0%
12
(
x
2
−
y
2
+
z
2
)
0%
12
(
x
2
+
y
2
+
z
2
)
Explanation
12
x
2
−
24
y
2
+
36
z
2
=
12
x
2
−
12
×
2
y
2
+
12
×
3
z
2
=
12
(
x
2
−
2
y
2
+
3
z
2
)
Factorisation of the expression :
7
a
2
−
49
b
2
Report Question
0%
7
(
a
2
−
7
b
2
)
0%
7
a
2
−
7
b
2
0%
7
(
a
2
−
b
2
)
0%
7
(
7
a
2
−
b
2
)
Explanation
7
a
2
−
49
b
2
=
(
7
×
a
×
a
)
−
(
7
×
7
×
b
×
b
)
=
7
(
a
2
−
7
b
2
)
Factorise:
80
p
2
−
72
p
Report Question
0%
8
p
(
5
p
−
9
)
0%
16
p
(
5
p
−
4
)
0%
8
p
(
10
p
−
8
)
0%
8
p
(
10
p
−
9
)
Explanation
80
p
2
−
72
p
=
(
8
×
10
×
p
×
p
)
−
(
8
×
9
×
p
)
=
8
p
(
10
p
−
9
)
Factorisation of the expression
17
p
2
q
−
102
p
q
2
results in:
Report Question
0%
17
p
q
(
p
−
8
q
)
0%
17
p
q
(
p
−
q
)
0%
17
p
q
(
p
−
4
q
)
0%
17
p
q
(
p
−
6
q
)
Explanation
17
p
2
q
−
102
p
q
2
=
(
17
×
p
×
p
×
q
)
−
(
17
×
6
×
p
×
q
×
q
)
=
17
p
q
(
p
−
6
q
)
The factorisation of
(
21
a
2
+
3
a
)
is
Report Question
0%
3
a
(
7
a
+
1
)
0%
7
a
(
3
a
+
1
)
0%
3
a
(
7
a
+
3
a
)
0%
3
a
(
a
+
7
)
Explanation
The factorisation of
21
a
2
+
3
a
is
3
a
(
7
a
+
1
)
Taking common terms out, we get
3
a
(
7
a
+
1
)
.
Multiplying factors is an example of
Report Question
0%
polynomial
0%
quadratic equation
0%
division algorithm
0%
factorisation
Explanation
Multiplying factors is an example of factorisation.
Example:
4
x
2
+
2
x
is a factor
2
x
(
2
x
+
1
)
By multiplying the factor we get
2
x
(
2
x
+
1
)
=
4
x
2
+
2
x
(
14
x
2
y
z
−
28
x
2
y
2
z
3
+
32
y
2
z
2
)
÷
(
−
4
x
y
)
is equal to
Report Question
0%
7
2
y
z
+
7
x
y
z
2
+
8
x
y
z
0%
−
7
2
x
z
+
7
x
y
z
3
−
8
y
z
2
x
0%
−
7
2
x
z
−
7
x
y
z
3
+
8
y
z
2
x
0%
7
2
x
z
−
7
x
y
z
2
−
8
y
z
2
x
Explanation
(
14
x
2
y
z
−
28
x
2
y
2
z
3
+
32
y
2
z
2
)
÷
(
−
4
x
y
)
=
14
x
2
y
z
−
4
x
y
−
28
x
2
y
2
z
3
−
4
x
y
+
32
y
2
z
2
−
4
x
y
=
−
7
2
x
z
+
7
x
y
z
3
−
8
y
z
2
x
Hence, B is correct.
4
a
+
12
b
is equal to
Report Question
0%
4
a
0%
12
b
0%
4
(
a
+
3
b
)
0%
3
a
Explanation
Given:
4
a
+
12
b
=
4
⋅
a
+
4
⋅
3
⋅
b
=
4
⋅
a
+
4
⋅
3
b
=
4
(
a
+
3
b
)
, (common factor
4
is taken out)
(
3
x
2
−
2
x
)
÷
x
Report Question
0%
3
x
0%
3
x
−
2
0%
3
x
+
2
0%
None
Explanation
Let us first factorize the given polynomial
3
x
2
−
2
x
by finding the HCF of all the terms as follows:
3
x
2
=
3
×
x
×
x
2
x
=
2
×
x
Therefore,
H
C
F
=
x
Now, we factor out the HCF from each term of the polynomial
3
x
2
−
2
x
as shown below:
3
x
2
−
2
x
=
x
(
3
x
−
2
)
Let us now divide the polynomial
3
x
2
−
2
x
by the monomial
x
:
3
x
2
−
2
x
x
=
x
(
3
x
−
2
)
x
=
3
x
−
2
Hence,
(
3
x
2
−
2
x
)
÷
x
=
3
x
−
2
.
(
5
a
3
b
−
7
a
b
3
)
÷
a
b
Report Question
0%
5
a
2
−
7
b
2
0%
5
a
−
7
b
2
0%
5
a
2
−
b
2
0%
None
Explanation
We divide the given polynomial
5
a
3
b
−
7
a
b
3
by the monomial
a
b
as shown below:
5
a
3
b
−
7
a
b
3
a
b
=
5
a
3
b
a
b
−
7
a
b
3
a
b
=
5
a
3
b
a
b
−
7
a
b
3
a
b
=
5
×
a
×
a
×
a
×
b
a
×
b
−
7
×
a
×
b
×
b
×
b
a
×
b
=
5
a
2
−
7
b
2
Hence,
5
a
3
b
−
7
a
b
3
a
b
=
5
a
2
−
7
b
2
.
(
4
l
5
−
6
l
4
+
8
l
3
)
÷
2
l
2
Report Question
0%
(
2
l
2
−
3
l
+
4
)
0%
l
(
2
l
2
−
3
l
+
4
)
0%
l
(
2
l
2
−
3
l
+
3
)
0%
None
Explanation
We divide the given polynomial
4
l
5
−
6
l
4
+
8
l
3
by the monomial
2
l
2
as shown below:
4
l
5
−
6
l
4
+
8
l
3
2
l
2
=
4
l
5
2
l
2
−
6
l
4
2
l
2
+
8
l
3
2
l
2
=
(
2
×
2
×
l
5
2
l
2
)
−
(
2
×
3
×
l
4
2
l
2
)
+
(
2
×
2
×
2
×
l
3
2
l
2
)
=
(
2
×
l
5
×
l
−
2
1
)
−
(
3
×
l
4
×
l
−
2
1
)
+
(
2
×
2
×
l
3
×
l
−
2
1
)
=
(
2
×
l
(
5
−
2
)
)
−
(
3
×
l
(
4
−
2
)
)
+
(
4
×
l
(
3
−
2
)
)
(
∵
a
x
+
a
y
=
a
x
+
y
)
=
2
l
3
−
3
l
2
+
4
l
=
l
(
2
l
2
−
3
l
+
4
)
Hence,
4
l
5
−
6
l
4
+
8
l
3
2
l
2
=
l
(
2
l
2
−
3
l
+
4
)
.
Factorize
5
x
y
z
2
−
3
z
y
Report Question
0%
y
z
(
5
x
z
−
3
)
0%
y
z
(
5
y
z
−
3
)
0%
x
z
(
5
x
z
−
3
y
)
0%
None of these
Explanation
We know that
a
(
b
+
c
)
=
a
b
+
a
c
and the reverse process
a
b
+
a
c
=
a
(
b
+
c
)
is called taking out the common factor.
Consider the factorization of the expression
5
x
y
z
2
−
3
z
y
:
5
x
y
z
2
=
5
×
x
×
y
×
z
×
z
3
z
y
=
3
×
z
×
y
⇒
H
C
F
(
5
x
y
z
2
,
3
z
y
)
=
z
×
y
=
y
z
Therefore, we have:
5
x
y
z
2
−
3
z
y
=
(
5
×
x
×
y
×
z
×
z
)
−
(
3
×
z
×
y
)
=
y
z
(
5
x
z
−
3
)
Hence,
5
x
y
z
2
−
3
z
y
=
y
z
(
5
x
z
−
3
)
.
15
(
a
3
b
2
c
2
−
a
2
b
3
c
2
+
a
2
b
2
c
3
)
÷
3
a
b
c
Report Question
0%
a
b
c
(
a
−
b
+
c
)
0%
5
a
b
c
(
a
+
b
+
c
)
0%
5
a
b
c
(
a
−
b
+
c
)
0%
None
Explanation
We divide the given polynomial
15
(
a
3
b
2
c
2
−
a
2
b
3
c
2
+
a
2
b
2
c
3
)
by the monomial
3
a
b
c
as shown below:
15
(
a
3
b
2
c
2
−
a
2
b
3
c
2
+
a
2
b
2
c
3
)
3
a
b
c
=
15
a
3
b
2
c
2
−
15
a
2
b
3
c
2
+
15
a
2
b
2
c
3
3
a
b
c
=
15
a
3
b
2
c
2
3
a
b
c
−
15
a
2
b
3
c
2
3
a
b
c
+
15
a
2
b
2
c
3
3
a
b
c
=
(
3
×
5
×
a
×
a
×
a
×
b
×
b
×
c
×
c
3
a
b
c
)
−
(
3
×
5
×
a
×
a
×
b
×
b
×
b
×
c
×
c
3
a
b
c
)
+
(
3
×
5
×
a
×
a
×
b
×
b
×
c
×
c
×
c
3
a
b
c
)
=
5
a
2
b
c
−
5
a
b
2
c
+
5
a
b
c
2
=
5
a
b
c
(
a
−
b
+
c
)
Hence,
15
(
a
3
b
2
c
2
−
a
2
b
3
c
2
+
a
2
b
2
c
3
)
3
a
b
c
=
5
a
b
c
(
a
−
b
+
c
)
.
(
3
p
3
−
9
p
2
q
−
6
p
q
2
)
÷
(
−
3
p
)
Report Question
0%
(
2
q
2
+
3
p
q
+
p
2
)
0%
(
2
q
2
+
p
q
−
p
2
)
0%
(
2
q
2
+
3
p
q
−
p
2
)
0%
None
Explanation
We divide the given polynomial
3
p
3
−
9
p
2
q
−
6
p
q
2
by the monomial
−
3
p
as shown below:
⇒
3
p
3
−
9
p
2
q
−
6
p
q
2
−
3
p
=
3
p
3
−
3
p
−
9
p
2
q
−
3
p
−
6
p
q
2
−
3
p
=
−
3
p
3
3
p
+
9
p
2
q
3
p
+
6
p
q
2
3
p
=
−
(
3
×
p
×
p
×
p
3
p
)
−
(
3
×
3
×
p
×
p
×
q
3
p
)
+
(
2
×
3
×
p
×
q
×
q
3
p
)
=
−
p
2
+
3
p
q
+
2
q
2
Hence,
3
p
3
−
9
p
2
q
−
6
p
q
2
−
3
p
=
2
q
2
+
3
p
q
−
p
2
.
Divide the given polynomial by the given monomial:
(
2
3
a
2
b
2
c
2
+
4
3
a
b
2
c
2
)
÷
1
2
a
b
c
Report Question
0%
4
3
(
a
b
c
+
b
c
)
0%
2
3
(
a
b
c
+
2
b
c
)
0%
4
3
(
a
b
c
+
2
b
c
)
0%
None
Explanation
We divide the given polynomial
2
3
a
2
b
2
c
2
+
4
3
a
b
2
c
2
by the monomial
1
2
a
b
c
as shown below:
2
3
a
2
b
2
c
2
+
4
3
a
b
2
c
2
1
2
a
b
c
=
2
3
a
2
b
2
c
2
1
2
a
b
c
+
4
3
a
b
2
c
2
1
2
a
b
c
=
2
3
×
a
×
a
×
b
×
b
×
c
×
c
1
2
×
a
×
b
×
c
+
4
3
×
a
×
b
×
b
×
c
×
c
1
2
×
a
×
b
×
c
=
(
2
3
×
2
×
a
×
b
×
c
)
+
(
4
3
×
2
×
b
×
c
)
=
4
3
a
b
c
+
8
3
b
c
=
4
3
(
a
b
c
+
2
b
c
)
Hence,
2
3
a
2
b
2
c
2
+
4
3
a
b
2
c
2
1
2
a
b
c
=
4
3
(
a
b
c
+
2
b
c
)
.
(
25
x
5
−
15
x
4
)
÷
5
x
3
Report Question
0%
x
(
5
x
+
3
)
0%
(
5
x
−
3
)
0%
x
(
5
x
−
3
)
0%
None
Explanation
We divide the given polynomial
25
x
5
−
15
x
4
by the monomial
5
x
3
as shown below:
25
x
5
−
15
x
4
5
x
3
=
25
x
5
5
x
3
−
15
x
4
5
x
3
=
(
5
×
5
×
x
5
5
x
3
)
−
(
3
×
5
×
x
4
5
x
3
)
=
(
5
×
x
5
×
x
−
3
1
)
−
(
3
×
x
4
×
x
−
3
1
)
=
(
5
×
x
(
5
−
3
)
)
−
(
3
×
x
(
4
−
3
)
)
(
∵
a
x
+
a
y
=
a
x
+
y
)
=
5
x
2
−
3
x
=
x
(
5
x
−
3
)
Hence,
25
x
5
−
15
x
4
5
x
3
=
x
(
5
x
−
3
)
.
Factorize the following expression
16
x
+
64
x
2
y
Report Question
0%
16
x
(
1
+
4
x
y
)
0%
6
x
(
1
+
4
x
y
)
0%
16
x
(
1
−
4
x
y
)
0%
None of these
Explanation
16
x
+
64
x
2
y
=
16
x
+
4
×
(
16
x
)
(
x
y
)
Taking
16
x
common
=
16
x
(
1
+
4
x
y
)
Factorize the following
p
q
−
p
r
−
3
p
s
Report Question
0%
p
(
q
+
r
−
3
s
)
0%
p
(
q
+
r
+
3
s
)
0%
q
(
q
+
r
−
3
s
)
0%
None of these
Explanation
p
q
−
p
r
−
3
p
s
(To factorise means to make factor)
Taking
p
common from all
3
equations, we have
p
(
q
−
r
−
3
s
)
Factorise the expression:
2
a
3
−
3
a
2
b
+
2
a
2
c
Report Question
0%
a
2
(
2
a
−
3
b
+
2
c
)
0%
a
2
(
2
a
+
3
b
+
2
c
)
0%
a
2
(
2
a
−
3
b
−
2
c
)
0%
None of these
Explanation
2
a
3
−
3
a
2
b
+
2
a
2
c
Taking
a
2
common from all
3
terms
=
a
2
(
2
a
−
3
b
+
2
c
)
Factorize the following expression:
10
x
3
−
25
x
4
y
Report Question
0%
5
x
3
(
2
−
5
x
y
)
0%
5
x
3
(
2
+
5
x
y
)
0%
x
3
(
2
−
5
x
y
)
0%
None of these
Explanation
Taking
5
x
3
common from all the terms
5
x
3
(
2
−
5
x
y
)
Find the quotient and remainder of the given expression,
(
3
x
3
+
4
x
2
−
5
)
÷
(
3
x
+
1
)
Report Question
0%
x
2
+
x
−
1
3
,
−
14
3
0%
x
2
−
3
x
+
5
9
,
14
3
0%
x
2
+
3
x
+
5
9
,
−
14
3
0%
None of these
The quotient when
x
3
−
5
x
2
+
7
x
−
4
is divided by
x
−
1
is.
Report Question
0%
x
2
+
4
x
+
3
0%
x
2
−
4
x
+
3
0%
x
2
−
4
x
−
3
0%
x
2
+
4
x
−
3
Find the quotient and remainder of the given expression,
(
3
x
3
+
2
x
2
+
7
x
−
5
)
÷
(
x
+
3
)
Report Question
0%
3
x
2
−
7
x
+
28
,
−
89
0%
3
x
2
−
7
x
+
40
,
125
0%
3
x
2
−
11
x
+
40
,
−
15
0%
None of these
Explanation
We divide
3
x
3
+
2
x
2
+
7
x
−
5
by
(
x
+
3
)
as shown in the above image:
From the division, we observe that the quotient is
3
x
2
−
7
x
+
28
and the remainder is
−
89
.
Hence,
the quotient is
3
x
2
−
7
x
+
28
and the remainder is
−
89
.
Find the quotient and remainder of the given expression,
(
2
x
4
−
7
x
3
−
13
x
2
+
63
x
−
48
)
÷
(
2
x
−
1
)
Report Question
0%
x
3
−
3
x
2
−
8
x
+
55
2
,
−
41
2
0%
x
3
−
3
x
2
−
8
x
+
55
2
,
41
2
0%
x
3
+
3
x
2
−
8
x
+
55
2
,
−
41
2
0%
None of these
Explanation
We divide
2
x
4
−
7
x
3
−
13
x
2
+
63
x
−
48
by
(
2
x
−
1
)
as shown in the above image:
From the division, we observe that the quotient is
x
3
−
3
x
2
−
8
x
+
55
2
and the remainder is
−
41
2
.
Hence,
the quotient is
x
3
−
3
x
2
−
8
x
+
55
2
and the remainder is
−
41
2
.
Find the quotient and remainder of the given expression,
(
8
x
4
−
2
x
2
+
6
x
−
5
)
÷
(
4
x
+
1
)
Report Question
0%
2
x
3
−
x
2
2
−
3
8
x
+
51
32
,
−
211
32
0%
2
x
3
−
x
2
2
−
3
8
x
+
51
32
,
211
32
0%
2
x
3
+
x
2
2
−
3
8
x
+
51
32
,
−
211
32
0%
None of these
The factorized form of
3x — 24
is
Report Question
0%
3x \times 24
0%
3(x - 8)
0%
24 (x - 3)
0%
3(x - 12)
Explanation
The given polynomial is
3x-24
Now,
3x-24
=3\times x-3\times 8
Taking out
3
as common,
= 3 (x – 8)
Hence,
Op-B
is correct.
Factorised form of
23xy - 46x + 54y - 108
is
Report Question
0%
(23x + 54) (y - 2)
0%
(23x + 54y) (y - 2)
0%
(23xy + 54y) (- 46x - 108)
0%
(23x + 54) (y + 2)
Explanation
We have
23xy - 46x + 54y - 108
Factoring out
23
from the first two terms and
54
from the last two,
\\=23x(y-2)+54(y-2)
Now, factoring out
(y-2),
= (23x+54)(y-2)
\therefore 23xy - 46x + 54y - 108 = (23x+54)(y-2)
Hence, option (A) is correct.
State whether the statement is true (T) or false (F).
Factorization of
-3a^{2} + 3ab + 3ac
is
3a(-a-b-c)
.
Report Question
0%
True
0%
False
Explanation
Given,
-3a^{2} + 3ab + 3ac
Factoring out
3a
, we get
=3a(-a+b+c)
\therefore -3a^{2} + 3ab + 3ac=3a(-a+b+c)
Hence, the given statement is false.
State whether the statement is true (T) or false (F).
Common factor of
12a^{2}b^{2} + 4ab^{2} - 32
is
4
.
Report Question
0%
True
0%
False
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Practice Class 8 Maths Quiz Questions and Answers
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