Question

Given that the sum of two of the roots of `x^3-3x^2-4x+12=0` is zero, how do you factor `x^3-3x^2-4x+12` ?

Answer 1

`x^3-3x^2-4x+12 = (x-2)(x+2)(x-3)`

Explanation:

If the roots are `alpha`, `beta` and `gamma` then:

`x^3-3x^2-4x+12 = (x-alpha)(x-beta)(x-gamma)`

`color(white)(x^3-3x^2-4x+12) = x^3-(alpha+beta+gamma)x^2+(alphabeta+betagamma+gammaalpha)x-alphabetagamma`

So, equating the coefficients of `x^2`, we find:

`alpha+beta+gamma = 3`

Since the sum of two of the roots is `0`, the remaining one must be `3`.

So `(x-3)` is a factor.

In fact we find:

`x^3-3x^2-4x+12 = x^2(x-3)-4(x-3)`

`color(white)(x^3-3x^2-4x+12) = (x^2-4)(x-3)`

`color(white)(x^3-3x^2-4x+12) = (x^2-2^2)(x-3)`

`color(white)(x^3-3x^2-4x+12) = (x-2)(x+2)(x-3)`