Question

Factorise :
$\left(i\right)ax+bx+ay+by$
$\left(ii\right)a{x}^2+b{y}^2+b{x}^2+a{y}^2$
$\left(iii\right)a^2+bc+ab+ac$
$\left(iv\right)ax–ay+bx–by$

Answer 1

We have,
(i) ax + bx + ay + by = (ax + bx) + (ay + by) [Grouping the terms]
= (a + b)x + (a + b)y
= (a + b) (x + y) [Taking (a + b) common]
$\left(ii\right)a{x}^2+b{y}^2+b{x}^2+a{y}^2=a{x}^2+b{x}^2+a{y}^2+b{y}^2$ [Re-arranging the terms]
$=(a+b)x^2+(a+b)y^2=(a+b)(x^2+y^2)$ [Taking (a + b) common]
$\left(iii\right)a^2+bc+ab+ac=(a^2+ab)+(ac+bc)$ [Re-grouping the terms]
$=a(a+b)+(a+b)c=(a+b)(a+c)$ [Taking (a + b) common]
$\left(iv\right)ax–ay+bx–by=a(x–y)+b(x–y)$
$=(a+b)(x–y)$ [Taking (x – y) common]