How do you factor x^3+3x^2-6x-8x3+3x26x8?

2 Answers
Oct 17, 2016

(x+1)(x-2)(x+4)(x+1)(x2)(x+4)

Explanation:

By trial and error
let f(x)=x^3+3x^2-6x-8f(x)=x3+3x26x8
let x=-1x=1
so f(-1)=-1+3+6-8=0f(1)=1+3+68=0
so (x+1)(x+1) is a factor
Then you have to make a long division
(x^3+3x^2-6x-8)/(x+1)=x^2+2x-8x3+3x26x8x+1=x2+2x8
Then factorise x^2+2x-8x2+2x8
x^2+2x-8=(x-2)(x+4)x2+2x8=(x2)(x+4)
and finally
x^3+3x^2-6x-8=(x+1)(x-2)(x+4)x3+3x26x8=(x+1)(x2)(x+4)

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

Oct 17, 2016

(x-2)(x+1)(x+4)(x2)(x+1)(x+4)

Explanation:

Group the terms in 'pairs' as follows.

[x^3-8]+[3x^2-6x][x38]+[3x26x]

Now, the first group is a color(blue)"difference of cubes"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)