In this page we have *Class 8 Maths Exercise 14.2 Solutions* for Factorisation.This exercise has questions on Factorisation using identities.

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Factorize the following expressions.

(i) a² + 8a + 16

(ii) p² – 10 p + 25

(iii) 25m² + 30m + 9

(iv) 49y² + 84yz + 36z²

(v) 4x² – 8x + 4

(vi) 121b² – 88bc + 16c²

(vii) (l + m) ² – 4lm

(viii) a

We have to make use of following identities to factorize them

(a+b)

(a-b)

a

(i) a² + 8a + 16

= a

So from first identity, it can be written as

=(a+4)

(ii) p² – 10 p + 25

= p

So from second identity, it can be written as

=(p-5)

(iii) 25m² + 30m + 9

= (5m)

So from first identity, it can be written as

=(5m+3)

(iv) 49y² + 84yz + 36z²

= (7y)

So from first identity, it can be written as

=(7y+6z)

(v) 4x² – 8x + 4

= (2x)

So from second identity, it can be written as

=(2x-2)

= 4(x-1)

(vi) 121b² – 88bc + 16c²

= (11b)

So from second identity, it can be written as

=(11b-4c)

(vii) (l + m) ² – 4lm

=l

= l

So from second identity, it can be written as

=(l-m)

(viii) a

= (a

So from first identity, it can be written as

=(a²+b²)²

Factorize.

(i) 4p² – 9q²

(ii) 63a² – 112b²

(iii) 49x² – 36

(iv) 16x

(v) (l + m) ² – (l – m) ²

(vi) 9x² y² – 16

(vii) (x² – 2xy + y²) – z²

(viii) 25a² – 4b² + 28bc – 49c²

We have to make use of following identities to factorize them

(a+b)

(a-b)

a

(i) 4p² – 9q²

So from third identity, it can be written as

=(2p-3q)( 2p+3q)

(ii) 63a² – 112b²

= 7( 9a

=7[ (3a)

So from third identity, it can be written as

=7(3a-4b)( 3a+4b)

(iii) 49x² – 36

=(7x)

So from third identity, it can be written as

=(7x-6)( 7x+6)

(iv) 16x

= x³(16x²-144)

= x³(4x+12)(4x-12)

(v) (l + m) ² – (l – m) ²

= [ l+m +l-m][l+m-l+m]

=2l×2m

=4lm

(vi) 9x² y² – 16

= (3xy-4)(3xy+4)

(vii) (x² – 2xy + y²) – z²

= (x-y)

Now as a

= (x-y+z)(x-y-z)

(viii) 25a² – 4b² + 28bc – 49c²

Factorizing each tem

= (5a)

Rearranging the terms

=(5a)

Now as (a-b)

=(5a)

=(5a-2b+7c)(5a+2b-7c)

Factorize the expressions.

(i) ax² + bx

(ii) 7p² + 21q²

(iii) 2x³ + 2xy² + 2xz²

(iv) am² + bm² + bn² + an²

(v) (lm + l) + m + 1

(vi) y (y + z) + 9 (y + z)

(vii) 5y² – 20y – 8z + 2yz

(viii) 10ab + 4a + 5b + 2

(ix) 6xy – 4y + 6 – 9x

(i) ax² + bx

=x(ax+b)

(ii) 7p² + 21q²

=7(p

(iii) 2x³ + 2xy² + 2xz²

=2x(x²+y²+z²)

(iv) am² + bm² + bn² + an²

Rearranging the terms

= am²+ an²+ bm² + bn²

=a(m

=(a+b) (m

(v) (lm + l) + m + 1

=l(m+1) +1(m+1)

=(l+1)(m+1)

(vi) y (y + z) + 9 (y + z)

= (y+z)(y+9)

(vii) 5y² – 20y – 8z + 2yz

=5y(y-4) +2z(y-4)

=(5y+2z)(y-4)

(viii) 10ab + 4a + 5b + 2

=2a(5b+2) +1(5b+2)

=(2a+1)(5b+2)

(ix) 6xy – 4y + 6 – 9x

=2y(3x-2) +3(2-3x)

=2y(3x-2)-3(3x-2)

=(2y-3)(3x-2)

Factorize.

(i) a

(ii) p

(iii) x

(iv) x

(v) a

(i) a

(ii) p

=(p²+9)(p²-9)

(iii) x

= (x²+(y+z) ²)(x²-(y+z) ²)

= (x²+(y+z) ²)[(x+y+z)(x-y-z)]

(iv) x

=(x²-(x-z) ²)(x²+(x-z) ²)

=[(x+x-z)(x-x+z)][(x²+(x-z) ²]

=z(2x-z) [(x²+(x-z) ²]

(v) a

=(a

Factorize the following expressions.

(i) p² + 6p + 8

(ii) q² – 10q + 21

(iii) p² + 6p – 16

(i) p²+6p+8

=p(p+6)+8

(ii) q²-10q+21

=q(q-10)+21

(iii) p²+6p-16

=p(p+6)-16

- Class 8 Maths Exercise 14.2 Solutions has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail. You can download the solutions as PDF in the below Link also

Download Factorisation Exercise 14.2 pdf - This chapter 14 has total 4 Exercise 14.1,14.2,14.3 and 14.4. This is the Second exercise in the chapter.You can explore previous exercise of this chapter by clicking the link below

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