Question

Factorise 25(2x+y) 2 -16(x-y) 2

Answer 1

25(2x+y)^2-16(x-y)^2
[5(2x+y)]^2-[4(x-y)]^2
using identity a^2-b^2=(a+b)(a-b)
[5(2x+y)+4(x-y)] [5(2x+y)-4(x-y)]
[10x+5y+4x-4y] [10x+5y-4x+4y]
(14x-y) (6x+9y)

Answer 2

The factorisation of $25(2x+y)^2-16(x-y)^2$ is $3(14x+y) (2x+3y)$.

Step-by-step explanation:

We have,

$25(2x+y)^2-16(x-y)^2$

To find, the factorisation of $25(2x+y)^2-16(x-y)^2$ = ?

∴ $25(2x+y)^2-16(x-y)^2$

$= [5(2x+y)]^2-[4(x-y)]^2$

$= [5(2x+y)+4(x-y)][5(2x-y)+4(x-y)]$

[ ∵$a^{2} -b^{2} =(a+b)(a-b)$]

$= (10x+5y+4x-4y) (10x+5y-4x+4y)$

$= (14x+y) (6x+9y)$

$= 3(14x+y) (2x+3y)$

Hence, the factorisation of $25(2x+y)^2-16(x-y)^2$ is $3(14x+y) (2x+3y)$.