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8th Grade > Mathematics

FACTORISATION MCQs

Total Questions : 57 | Page 1 of 6 pages
Question 1. Which of the following is the expression obtained by the division of 5pq(p2q2) by 2p(p+q)?
  1.    5q(p−q)4
  2.    5q(p+q)2
  3.    5q(p−q)2
  4.    5q(p+q)4
 Discuss Question
Answer: Option C. -> 5q(p−q)2
:
C
Given, the expression is
5pq(p2q2).
[Using the identity:a2b2=(a+b)(ab)]
5pq(p2q2)=5pq(p+q)(pq)
Nowdividing the above expression by
2p(p+q),we get
5pq(p+q)(pq)2p(p+q)
=5q(pq)2
Question 2. Division of 72x2y4z6 by 8x2y2z3 gives 9y2z3
  1.    True
  2.    False
  3.    5q(p−q)2
  4.    5q(p+q)4
 Discuss Question
Answer: Option A. -> True
:
A
728=9y4y2=y2z6z3=z3
Therefore,72x2y4z68x2y2z3=9y2z3
Question 3. Solve:4yz(z2+6z16)÷2y(z+8) 
  1.    z−2
  2.    z(z−2)
  3.    2z(z−2)
  4.    z(z−4)
 Discuss Question
Answer: Option C. -> 2z(z−2)
:
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)=6andab=16.
Since,
8+(2)=6and(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)
Question 4. Divide m2+7m60 by (m5).
  1.    4(m - 6)
  2.    2(m + 6)
  3.    (m + 12)
  4.    (m - 12)
 Discuss Question
Answer: Option C. -> (m + 12)
:
C
Given, the expression is
m2+7m60(m5).
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+7m60=m2+12m5m60=m(m+12)5(m+12)=(m5)(m+12)
Therefore,
m2+7m60(m5)=(m5)(m+12)(m5)=(m+12)
Question 5. On dividing (24xy2)(z2) by 6yz2, we get
  1.    4xz
  2.    4xy
  3.    4yz
  4.    4xyz
 Discuss Question
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.
Question 6. Find the common factor between
2t,3t2and 4.
  1.    3
  2.    2
  3.    1
  4.    t
 Discuss Question
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.
Question 7. Find the value of the given expression
(x3+2x2+3x)÷2x
  1.    (x2+2x+3)2
  2.    12
  3.    (x2+2x+3)​
  4.    1
 Discuss Question
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
Question 8. The factors of  ax + bx - ay - by are __________.
  1.    (a + b) and (x + y)
  2.    (a - b) and (x - y)
  3.    (a + b) and (x - y)
  4.    (a - b) and (x + y)
 Discuss Question
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
Question 9. Which among the following is not a factor of  4y(y+3)?
  1.    y2 + 3y
  2.    4y2
  3.    4(y + 3)
  4.    4y
 Discuss Question
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.
Question 10. Substituting x=3 in (x2)5x gives 
___
 Discuss Question

:
Substituting x=3 in x25x gives
(3)25(3)=9+15=24

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