How do you factor #x^3 + 6x^2 - 9x - 54# by grouping?

1 Answer
Jul 23, 2016

Answer:

(x+6)(x-3)(x+3)

Explanation:

Group the terms into 'pairs' as follows.

#[x^3+6x^2]+[-9x-54]#

Now, factorise each 'pair'

#color(red)(x^2)(x+6)color(red)(-9)(x+6)#

Take out the common factor of (x +6)

#(x+6)color(red)((x^2-9))#

#color(red)(x^2-9)" is a "color(blue)"difference of squares"# and in general factorises.

#color(red)(|bar(ul(color(white)(a/a)color(black)(a^2-b^2=(a-b)(a+b))color(white)(a/a)|)))#

#x^2=(x)^2" and " 9=(3)^2rArra=x" and " b=3#

#rArrcolor(red)(x^2-9)=(x-3)(x+3)#

Pulling this altogether we get.

#x^3+6x^2-9x-54=(x+6)(x-3)(x+3)#