NCERT Solutions Class 8 Mathematics Solutions for Exercise 14.2 in Chapter 14 - Factorisation

Question 19 Exercise 14.2

Q3) Factorise the expressions:

(i) ax^2+bx

(ii) 7p^2+21q^2

(iii) 2x^3+2xy^2+2xz^2

(iv) am^2+bm^2+bn^2+an^2

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

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

(vii) 5y^2-20y-8z+2yz

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




(i) x(ax+b)

(ii) 7\left(p^2+3q^2\right)

(iii) 2x\left(x^2+y^2+z^2\right)

(iv) Lets split the expression to am^2+bm^2and\ an^2+bn^2

Therefore, am^2+bm^{2\ }

=\ m^2\left(a+b\right)

and, an^2+bn^2


On adding these two

m^2\left(a+b\right)\ +n^2\left(a+b\right)

\left(a+b\right)\ \left(m^2+n^2\right)

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



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

(vii) Let's split the expression to \left(5y^2-20y\right)\ and\ \left(2yz-8z\right)

= 5y (y-4)

and, 2z(y-4)

on adding the two

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


(viii) Let's the expression to 10ab +4a and 5b+2

For, 10 ab +4a

= 2a(5b+2)

For, 5b+2


now adding the two,

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

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

(ix) Let's split the expression to 6xy-4y and 6-9x

For, 6xy-4y

= 2y (3x-2)

For, 6-9x


on adding the two expression

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

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

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