Free Factorisation 01 Practice Test - 8th Grade
Question 1
Which of the following is the factorised form of 20l2m+30alm?
SOLUTION
Solution : B
Given, the expression is 20l2m+30alm.
The given expression can be factorised as follows :-
20l2m+30alm
=(2×2×5×l×l×m)+(2×3×5×a×l×m)
Taking out the common factors from both the terms, we get
=(2×5×l×m)[(2×l)+(3×a)]
=10lm(2l+3a)
Question 2
Solve:4yz(z2+6z−16)÷2y(z+8)
z−2
z(z−2)
2z(z−2)
z(z−4)
SOLUTION
Solution : 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)=6 and ab=−16.
Since,
8+(−2)=6 and (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)
Question 3
Dividing x(3x2−27) by 3(x−3) gives x2+3x.
True
False
SOLUTION
Solution : A
The given expression is
x(3x2−27)
Taking 3 common we get
= 3x(x2−9)
[Using a2−b2=(a+b)(a−b)]
= 3x(x+3)(x−3)
= 3x(x+3)(x−3)3(x−3)
= x(x+3)
= x2+3x
Hence, the given statement is true
Question 4
Dividing (a+b)2+(a−b)2 by (a2+b2) gives
SOLUTION
Solution :(a+b)2+(a−b)2=(a2+b2+2ab)+(a2+b2−2ab)=2(a2+b2)
Dividing by a2+b2 we get,
= 2(a2+b2)(a2+b2)=2
Thus, ((a+b)2+(a−b)2)(a2+b2)=2
Question 5
Dividing 2x2+13x+15 by x+5 and then substituting x=2 in the resulting expression gives
SOLUTION
Solution :2x2+13x+15=2x2+10x+3x+15=2x(x+5)+3(x+5)=(2x+3)(x+5)(x+5)=2x+3
Substituting x=2 in resulting expression gives 2×2+3=7.
Question 6
Which of the following is/are the factor(s)
of 4x3+12x2+5x+15?
(2x2+5)
(4x2+5)
(x+3)
(2x2+3)
SOLUTION
Solution : B and C
Given, the expression is
4x3+12x2+5x+15.
Taking 4x2 common from the first two-terms and 5 common from the last two-terms, we get
4x2(x+3)+5(x+3)
=(x+3)(4x2+5)
Thus, (x+3) and (4x2+5)
are two factors of the given expression.
Question 7
A factor of a2−2ab+b2−c2 is ___________.
a−b
a−b−c
a+b+c
a+b−c
SOLUTION
Solution : B
a2−2ab+b2−c2=(a−b)2–c2
Using identity x2−y2=(x+y)(x−y) we get
(a−b)2–c2 = [(a−b)+c][(a−b)−c]
Hence, factors of a2−2ab+b2−c2 are a−b+c and a−b−c
Question 8
Divide (a2+7a+10) by (a+5).
SOLUTION
Solution : D
Given: a2+7a+10(a+5) ...(i)
Comparing a2+7a+10 with the identity x2+(a+b)x+ab,
we note that,(a+b)=7 and ab=10
So, 5+2=7 and (5)(2)=10
Hence,
a2+7a+10
=a2+5a+2a+10
=a(a+5)+2(a+5)
=(a+2)(a+5)
From (i), we get
a2+7a+10(a+5)=(a+2)(a+5)(a+5)
=(a+2)
Question 9
Divide m2+7m−60 by (m−5).
4(m - 6)
2(m + 6)
(m + 12)
(m - 12)
SOLUTION
Solution : C
Given, the expression is
m2+7m−60(m−5).
Now, comparing the numerator with the identity x2+(a+b)x+ab
We note that,
(a+b)=7 and ab=−60.
So,
12+(−5)=7 and (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)
Question 10
Factorise: 25+70x+49x2
(5x+7)(5x+2)
(7x+5)(5x+3)
(7x+5)(7x+5)
(3x+7)2
SOLUTION
Solution : C
Given, the expression is 25+70x+49x2.
The given expression can be rewritten as,
49x2+70x+25
Using the identity: (a+b)2=a2+2ab+b2,we get
49x2+70x+25=(7x)2+2(5)(7x)+52
=(7x+5)2
=(7x+5)(7x+5)
∴25+70x+49x2=(7x+5)(7x+5)