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

Jan 23, 2017

The zeros of $g \left(x\right)$ are $2$, $- 2$ and $- 3$

#### Explanation:

The difference of squares identity can be written:

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

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

Given:

$g \left(x\right) = {x}^{3} + 3 {x}^{2} - 4 x - 12$

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:

${x}^{3} + 3 {x}^{2} - 4 x - 12 = \left({x}^{3} + 3 {x}^{2}\right) - \left(4 x + 12\right)$

$\textcolor{w h i t e}{{x}^{3} + 3 {x}^{2} - 4 x - 12} = {x}^{2} \left(x + 3\right) - 4 \left(x + 3\right)$

$\textcolor{w h i t e}{{x}^{3} + 3 {x}^{2} - 4 x - 12} = \left({x}^{2} - 4\right) \left(x + 3\right)$

$\textcolor{w h i t e}{{x}^{3} + 3 {x}^{2} - 4 x - 12} = \left({x}^{2} - {2}^{2}\right) \left(x + 3\right)$

$\textcolor{w h i t e}{{x}^{3} + 3 {x}^{2} - 4 x - 12} = \left(x - 2\right) \left(x + 2\right) \left(x + 3\right)$

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