8th Grade > Mathematics
FACTORISATION MCQs
Total Questions : 57
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Answer: Option C. -> 5q(p−q)2
:
C
Given, the expression is
5pq(p2−q2).
[Using the identity:a2−b2=(a+b)(a−b)]
⇒5pq(p2−q2)=5pq(p+q)(p−q)
Nowdividing the above expression by
2p(p+q),we get
5pq(p+q)(p−q)2p(p+q)
=5q(p−q)2
:
C
Given, the expression is
5pq(p2−q2).
[Using the identity:a2−b2=(a+b)(a−b)]
⇒5pq(p2−q2)=5pq(p+q)(p−q)
Nowdividing the above expression by
2p(p+q),we get
5pq(p+q)(p−q)2p(p+q)
=5q(p−q)2
Answer: Option A. -> True
:
A
728=9y4y2=y2z6z3=z3
Therefore,72x2y4z68x2y2z3=9y2z3
:
A
728=9y4y2=y2z6z3=z3
Therefore,72x2y4z68x2y2z3=9y2z3
Answer: Option C. -> 2z(z−2)
:
C
Given, the expression is
4yz(z2+6z−16)2y(z+8).
First, we will factorise (z2+6z−16).
Comparing the above expression with the identity x2+(a+b)x+ab, we note that,(a+b)=6andab=−16.
Since,
8+(−2)=6and(8)(−2)=−16,
the expression, z2+6z−16
=z2+8z−2z−16
=z(z+8)−2(z+8)
=(z−2)(z+8) . . . (i)
By substituting (i) in the given expression, we get
=4yz(z−2)(z+8)2y(z+8)
=4×y×z×(z−2)×(z+8)2×y×(z+8)
=2z(z−2)
:
C
Given, the expression is
4yz(z2+6z−16)2y(z+8).
First, we will factorise (z2+6z−16).
Comparing the above expression with the identity x2+(a+b)x+ab, we note that,(a+b)=6andab=−16.
Since,
8+(−2)=6and(8)(−2)=−16,
the expression, z2+6z−16
=z2+8z−2z−16
=z(z+8)−2(z+8)
=(z−2)(z+8) . . . (i)
By substituting (i) in the given expression, we get
=4yz(z−2)(z+8)2y(z+8)
=4×y×z×(z−2)×(z+8)2×y×(z+8)
=2z(z−2)
Answer: Option C. -> (m + 12)
:
C
Given, the expression is
m2+7m−60(m−5).
Now, comparing the numeratorwith the identity x2+(a+b)x+ab
We note that,
(a+b)=7andab=−60.
So,
12+(−5)=7and(12)(−5)=−60.
Hence,
m2+7m−60=m2+12m–5m–60=m(m+12)–5(m+12)=(m−5)(m+12)
Therefore,
m2+7m−60(m−5)=(m−5)(m+12)(m−5)=(m+12)
:
C
Given, the expression is
m2+7m−60(m−5).
Now, comparing the numeratorwith the identity x2+(a+b)x+ab
We note that,
(a+b)=7andab=−60.
So,
12+(−5)=7and(12)(−5)=−60.
Hence,
m2+7m−60=m2+12m–5m–60=m(m+12)–5(m+12)=(m−5)(m+12)
Therefore,
m2+7m−60(m−5)=(m−5)(m+12)(m−5)=(m+12)
Answer: Option B. -> 4xy
:
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.
:
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.
Answer: Option C. -> 1
:
C
2t,3t2and4can 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.
:
C
2t,3t2and4can 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.
Answer: Option A. -> (x2+2x+3)2
:
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
:
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
Answer: Option C. -> (a + b) and (x - y)
:
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 ofax + bx - ay - by
:
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 ofax + bx - ay - by
Answer: Option B. -> 4y2
:
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, 4y2is a term in the expansion of 4y(y+3). It isnot a factor.
:
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, 4y2is a term in the expansion of 4y(y+3). It isnot a factor.
:
Substituting x=−3 in x2−5x gives
(−3)2−5(−3)=9+15=24