Free Factorisation 02 Practice Test - 8th Grade
Question 1
Which of the following is/are factor(s) of
49p2−36?
SOLUTION
Solution : A and B
The given expression can be written as,
49p2−36=(7p)2−(6)2
[Using the identity: a2−b2=(a+b)(a−b)]
=(7p+6)(7p−6)
Hence, (7p+6) and (7p−6)
are the factors of 49p2−36.
Question 2
On dividing (24xy2)(z2) by 6yz2, we get
4xz
4xy
4yz
4xyz
SOLUTION
Solution : 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 3
If 3 is a factor of 14pq, then 3 is also a factor of 28p2q.
True
False
SOLUTION
Solution : A
14pq itself is a factor of 28p2q. Thus, any factor of 14pq will also be a factor of 28p2q.
Question 4
Find the value of the given expression
(x3+2x2+3x)÷2x
(x2+2x+3)2
12
(x2+2x+3)
1
SOLUTION
Solution : 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 5
The value of x(x+1)(x+2)(x+3)x(x+1) is __________.
(x+2)
(x + 2) (x + 3)
(x+3)
x(x+1)
SOLUTION
Solution : B
Given,
x(x+1)(x+2)(x+3)x(x+1)
On cancelling the common terms x and (x+1) , we get
⇒(x+2)(x+3)
Question 6
The factors of ax + bx - ay - by are __________.
(a + b) and (x + y)
(a - b) and (x - y)
(a + b) and (x - y)
(a - b) and (x + y)
SOLUTION
Solution : 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 of ax + bx - ay - by
Question 7
Find the common factor between
2t,3t2and 4.
3
2
1
t
SOLUTION
Solution : C
2t, 3t2 and 4 can 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 8
Factorise:4y2−12y+9
(7y−5)(7y−5)
(5y−3)(5y−3)
(2y−3)(2y−3)
(2y−5)(2y−5)
SOLUTION
Solution : C
The given expression can be rewritten as,
(2y)2−2(2y)(3)+32.
Comparing with the identity
(a−b)2=a2−2ab+b2
we get,
(2y)2−2(2y)(3)+32
=(2y−3)2
=(2y−3)(2y−3)
Question 9
Which among the following is not a factor of 4y(y+3)?
y2 + 3y
4y2
4(y + 3)
4y
SOLUTION
Solution : 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, 4y2 is a term in the expansion of 4y(y+3). It is not a factor.
Question 10
4x2 is a factor of 4x(x+3)
True
False
SOLUTION
Solution : B
4x(x+3) = 4x2 + 3
So, 4x2 is a term of a given expression. 4x2 + 3 is not divisible by 4x2.
So, 4x2 is not a factor of 4x2+3.
The factors of 4x(x+3) are
4,x and (x+3)