# NCERT Solutions Class 8 Mathematics Solutions for Exercise 14.1 in Chapter 14 - Factorisation

Question 6 Exercise 14.1

Q3) Factorize :

(i) x^2+xy+8x+8y

(ii) 15xy-6x+5y-2

(iii) ax+bx-ay-by

(iv) 15pq+15+9q+25p

(v) z-7+7xy-xyz

Answer:

Solution3:

(i) Let's break this, into x^2+xy\ and\ 8x+8y

x^2+xy\ =\ x\times x+x\times y

8x+8y = 8\times x+8\times y

on summing factors of both

we get, x(x+y) + 8(x+y)

= (x+y) (x+8)

(ii) Let's break this, into 15xy+6x and [5y-2]

15xy - 6x = 3\times5\times x\times y-3\times2\times x

3\times x\left(5\times y-2\right)

3x\left(5y-2\right)

5y-2 = = 5\times y-2\times1

= 1(5y-2)

on adding the two,

Therefore, 3x (5y-2) + 1(5y-2)

= (3x-1)(5y-2)

(iii) Let's break this, into ax+bx and (-) [ay+by]

ax+bx= a\times x\ +\ b\times x

= x(a+b)

-ay-by = \left(-1\right)\times a\times y\ +\ \left(-1\right)\times b\times y

\left(-1\right)\times y\left[a+b\right]

on adding the two,

x(a+b) -y (a+b)

= (a+b)(x-y)

(iv) On breaking this into 15pq+9q and 15+25p

15pq+9q = 3\times5\times p\times q\ +\ 3\times3\times q

3\times q\ \left(5\times p+3\right)

on joining the two,

3q(5p+3) + 5(5p+3)

= (3q+5)(5p+3)

(v) Let's break this into z-7 and 7xy - xyz

z-7 = z\times1+7\times\left(-1\right)

\left(-1\right)\ \left[z\times\left(-1\right)+7\right]

\left(7-z\right)\left(-1\right)

7xy - xyz = 7\times x\times y\ -\ z\times x\times y

x\times y\ \times\left(7-z\right)

xy\ \left(7-z\right)

on adding the two

= xy (7-z) + (-1) (7-z)

= (7-z) (xy-1)

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