Question

How do you find all zeros of `g(x)=x^3+3x^2-4x-12`?

Answer 1

The zeros of `g(x)` are `2`, `-2` and `-3`

Explanation:

The difference of squares identity can be written:

`a^2-b^2 = (a-b)(a+b)`

We will use this with `a=x` and `b=2`.

Given:

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

Note that the ratio between the first and second terms is the same as that between the third and fourth terms.

So this cubic will factor by grouping:

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

`color(white)(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)`

Hence zeros: `x=2`, `x=-2`, `x=-3`.