# Given that the sum of two of the roots of x^3-3x^2-4x+12=0 is zero, how do you factor x^3-3x^2-4x+12 ?

Mar 12, 2017

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

#### Explanation:

If the roots are $\alpha$, $\beta$ and $\gamma$ then:

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

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

So, equating the coefficients of ${x}^{2}$, we find:

$\alpha + \beta + \gamma = 3$

Since the sum of two of the roots is $0$, the remaining one must be $3$.

So $\left(x - 3\right)$ is a factor.

In fact we find:

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

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

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

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