# 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?

Nov 1, 2016

The zeros of $f \left(x\right)$ are:

$x = \pm \sqrt{6} \text{ }$ (irrational)

$x = 2 \text{ }$ (rational)

#### Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

We use this with $a = x$ and $b = \sqrt{6}$ later.

Given:

$f \left(x\right) = {x}^{3} - 2 {x}^{2} - 6 x + 12$

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:

${x}^{3} - 2 {x}^{2} - 6 x + 12 = \left({x}^{3} - 2 {x}^{2}\right) - \left(6 x - 12\right)$

$\textcolor{w h i t e}{{x}^{3} - 2 {x}^{2} - 6 x + 12} = {x}^{2} \left(x - 2\right) - 6 \left(x - 2\right)$

$\textcolor{w h i t e}{{x}^{3} - 2 {x}^{2} - 6 x + 12} = \left({x}^{2} - 6\right) \left(x - 2\right)$

$\textcolor{w h i t e}{{x}^{3} - 2 {x}^{2} - 6 x + 12} = \left({x}^{2} - {\left(\sqrt{6}\right)}^{2}\right) \left(x - 2\right)$

$\textcolor{w h i t e}{{x}^{3} - 2 {x}^{2} - 6 x + 12} = \left(x - \sqrt{6}\right) \left(x + \sqrt{6}\right) \left(x - 2\right)$

Hence the zeros of $f \left(x\right)$ are:

$x = \pm \sqrt{6} \text{ }$ (irrational)

$x = 2 \text{ }$ (rational)