How do you factor #x^3+3x^2-6x-8#?

2 Answers
Oct 17, 2016

Answer:

#(x+1)(x-2)(x+4)#

Explanation:

By trial and error
let #f(x)=x^3+3x^2-6x-8#
let #x=-1#
so #f(-1)=-1+3+6-8=0#
so #(x+1)# is a factor
Then you have to make a long division
#(x^3+3x^2-6x-8)/(x+1)=x^2+2x-8#
Then factorise #x^2+2x-8#
#x^2+2x-8=(x-2)(x+4)#
and finally
#x^3+3x^2-6x-8=(x+1)(x-2)(x+4)#

graph{x^3+3x^2-6x-8 [-10, 10, -5, 5]}

Oct 17, 2016

Answer:

#(x-2)(x+1)(x+4)#

Explanation:

Group the terms in 'pairs' as follows.

#[x^3-8]+[3x^2-6x]#

Now, the first group is a #color(blue)"difference of cubes"# and factorises, in general, as.

#color(red)(bar(ul(|color(white)(2/2)color(black)(a^3-b^3=(a-b)(a^2+ab+b^2))color(white)(2/2)|)))#

Now #(x)^3=x^3" and " (2)^3=8#

#rArra=x" and " b=2#

#x^3-8=(x-2)(x^2+2x+2^2)=(x-2)(x^2+2x+4)#

The second group has a #color(blue)"common factor"# of 3x.

#rArr3x^2-6x=3x(x-2), "hence"#

#x^3-8+3x^2-6x=(x-2)(x^2+2x+4)+3x(x-2)#

There is now a #color(blue)"common factor " (x -2)#

#color(red)((x-2))(color(magenta)(x^2+2x+4+3x))=(x-2)(x^2+5x+4)#

#=(x-2)(x+1)(x+4)#