Question

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

Answer 1

Factor by grouping to find zeros: `2, -2, -3`

Explanation:

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

So this cubic factors by grouping:

`x^3+3x^2-4x-12`

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

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

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

`=(x-2)(x+2)(x+3)`

Hence zeros: `2, -2, -3`

Answer 2

`(x+2)(x-2)(x+3)`

Explanation:

We essentially have two parts to our function:

`color(steelblue)(x^3+3x^2)color(purple)(-4x-12)`

Let's factor a `x^2` out of the blue term, and a `-4` out of the purple term. We now have

`color(steelblue)(x^2(x+3))color(purple)(-4(x+3))`

Since both terms have an `x+3` in common, we can factor that out. We now have

`color(red)((x^2-4))(x+3)`

You might immediately recognize that what I have in red is a difference of squares, which can be factored as `(x+2)(x-2)`.

We now have

`(x+2)(x-2)(x+3)`

Hope this helps!