Question

factorization of [x^2 +(a+b+c)x+ab+bc] is -

Answer 1

x²+ax+bx+cx+ab+bc
x(x+b)+a(x+b)+c(x+b)
(x+b)(x+a+c)

Answer 2

The factorisation of $x^2+(a+b+c)x+ab+bc$ is equal to $(x+b)(x+a+c).$

Step-by-step explanation:

We have,

$x^2+(a+b+c)x+ab+bc$

To find, the factorisation of $x^2+(a+b+c)x+ab+bc=?$

∴ $x^2+(a+b+c)x+ab+bc$

$=x^2+ax+bx+cx+ab+bc$

$=(x^2+bx)+(ax+ab)+(cx+bc)$

$=x(x+b)+a(x+b)+c(x+b)$

Taking (x + b) as common, we get

$=(x+b)(x+a+c)$

The factorisation of $x^2+(a+b+c)x+ab+bc=(x+b)(x+a+c)$.

Hence, the factorisation of $x^2+(a+b+c)x+ab+bc$ is equal to $(x+b)(x+a+c).$