Question

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

Answer 1

`(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]}

Answer 2

`(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)`