Question

factorise x²+(a+b+c)x+ab+bc​

Answer 1

Answer:

The factorisation of x^2+(a+b+c)x+ab+bcx

2

+(a+b+c)x+ab+bc is equal to (x+b)(x+a+c).(x+b)(x+a+c).

step by step explanation:

We have,

x^2+(a+b+c)x+ab+bcx

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+(a+b+c)x+ab+bcx

2

+(a+b+c)x+ab+bc

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

2

+ax+bx+cx+ab+bc

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

2

+bx)+(ax+ab)+(cx+bc)

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

Taking (x + b) as common, we get

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

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

2

+(a+b+c)x+ab+bc=(x+b)(x+a+c) .

Hence, the factorisation of x^2+(a+b+c)x+ab+bcx

2

+(a+b+c)x+ab+bc is equal to (x+b)(x+a+c).(x+b)(x+a+c).