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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

1) a² + 8a + 16

= a

So from first identity, it can be written as

=(a+4)

2) p² – 10 p + 25

= p

So from second identity, it can be written as

=(p-5)

3) 25m² + 30m + 9

= (5m)

So from first identity, it can be written as

=(5m+3)

4) 49y² + 84yz + 36z²

= (7y)

So from first identity, it can be written as

=(7y+6z)

5) 4x² – 8x + 4

= (2x)

So from second identity, it can be written as

=(2x-2)

= 4(x-1)

6) 121b² – 88bc + 16c²

= (11b)

So from second identity, it can be written as

=(11b-4c)

7) (l + m) ² – 4lm

=l

= l

So from second identity, it can be written as

=(l-m)

8) 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

1) 4p² – 9q²

So from third identity, it can be written as

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

2) 63a² – 112b²

= 7( 9a

=7[ (3a)

So from third identity, it can be written as

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

3) 49x² – 36

=(7x)

So from third identity, it can be written as

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

4) 16x

= x³(16x²-144)

= x³(4x+12)(4x-12)

5) (l + m) ² – (l – m) ²

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

=2l×2m

=4lm

6) 9x² y² – 16

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

7) (x² – 2xy + y²) – z²

= (x-y)

Now as a

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

8) 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

1) ax² + bx

=x(ax+b)

2) 7p² + 21q²

=7(p

3) 2x³ + 2xy² + 2xz²

=2x(x²+y²+z²)

4) am² + bm² + bn² + an²

Rearranging the terms

= am²+ an²+ bm² + bn²

=a(m

=(a+b) (m

5) (lm + l) + m + 1

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

=(l+1)(m+1)

6) y (y + z) + 9 (y + z)

= (y+z)(y+9)

7) 5y² – 20y – 8z + 2yz

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

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

8) 10ab + 4a + 5b + 2

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

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

9) 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

1) p²+6p+8

=p(p+6)+8

2) q²-10q+21

=q(q-10)+21

3) p²+6p-16

=p(p+6)-16

Download Factorisation Exercise 14.2 pdf

**Notes**-
**Worksheets** -
**Ncert Solutions**

Class 8 Maths Class 8 Science

Given below are the links of some of the reference books for class 8 Math.

- Mathematics Foundation Course for JEE/Olympiad : Class 8 This book can take students maths skills further. Only buy if child is interested in Olympiad/JEE foundation courses.
- Mathematics for Class 8 by R S Aggarwal Detailed Mathematics book to clear basics and concepts. I would say it is a must have book for class 8 student.
- Pearson Foundation Series (IIT -JEE / NEET) Physics, Chemistry, Maths & Biology for Class 8 (Main Books) | PCMB Combo : These set of books could help your child if he aims to get extra knowledge of science and maths. These would be helpful if child wants to prepare for competitive exams like JEE/NEET. Only buy if you can provide help to the child while studying.
- Reasoning Olympiad Workbook - Class 8 :- Reasoning helps sharpen the mind of child. I would recommend students practicing reasoning even though they are not appearing for Olympiad.

You can use above books for extra knowledge and practicing different questions.

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