How do you factor #x^3 + x^2 -9x -9# by grouping?
1 Answer
Jun 30, 2016
(x+1)(x-3)(x+3)
Explanation:
Group the terms into 'pairs' as follows.
#[x^3+x^2]+[-9x-9]# now factorise each pair.
#color(red)(x^2)(x+1)-color(red)(9)(x+1)# We now have a common factor of (x+1) which can be 'taken out'
#(x+1)(color(red)(x^2-9))........ (A)#
#color(red)(x^2-9)" is a " color(blue)"difference of squares"# and is factorised in general as follows.
#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#
#rArrx^2-9=(x-3)(x+3)# Substitute these factors into (A)
#rArr(x+1)(x^2-9)=(x+1)(x-3)(x+3)# Thus
#x^3+x^2-9x-9=(x+1)(x-3)(x+3)#