Question

How do you find all of the real zeros of `f(x)=x^3-2x^2-6x+12` and identify each zero as rational or irrational?

Answer 1

The zeros of `f(x)` are:

`x = +-sqrt(6)" "` (irrational)

`x = 2" "` (rational)

Explanation:

The difference of squares identity can be written:

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

We use this with `a=x` and `b=sqrt(6)` later.

Given:

`f(x) = x^3-2x^2-6x+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^3-2x^2-6x+12 = (x^3-2x^2)-(6x-12)`

`color(white)(x^3-2x^2-6x+12) = x^2(x-2)-6(x-2)`

`color(white)(x^3-2x^2-6x+12) = (x^2-6)(x-2)`

`color(white)(x^3-2x^2-6x+12) = (x^2-(sqrt(6))^2)(x-2)`

`color(white)(x^3-2x^2-6x+12) = (x-sqrt(6))(x+sqrt(6))(x-2)`

Hence the zeros of `f(x)` are:

`x = +-sqrt(6)" "` (irrational)

`x = 2" "` (rational)