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