# How do you factor y= x^3+x^2+2x-4 ?

Mar 20, 2016

${x}^{3} + {x}^{2} + 2 x - 4$

$= \left(x - 1\right) \left({x}^{2} + 2 x + 4\right)$

$= \left(x - 1\right) \left(x + 1 - \sqrt{3} i\right) \left(x + 1 + \sqrt{3} i\right)$

#### Explanation:

First note that the sum of the coefficients is zero.

That is $1 + 1 + 2 - 4 = 0$.

So $x = 1$ is a zero and $\left(x - 1\right)$ a factor:

${x}^{3} + {x}^{2} + 2 x - 4 = \left(x - 1\right) \left({x}^{2} + 2 x + 4\right)$

The remaining quadratic factor is of the form $a {x}^{2} + b x + c$ with $a = 1$, $b = 2$ and $c = 4$.

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {2}^{2} - \left(4 \cdot 1 \cdot 4\right) = - 12$

Since this is negative, the quadratic has no linear factors with Real coefficients.

If we want to proceed further, we can use the quadratic formula to find the Complex roots and derive factors:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \frac{- b \pm \sqrt{\Delta}}{2 a}$

$= \frac{- 2 \pm \sqrt{- 12}}{2}$

$= \frac{- 2 \pm \sqrt{12} i}{2}$

$= \frac{- 2 \pm 2 \sqrt{3} i}{2}$

$= - 1 \pm \sqrt{3} i$

Hence:

${x}^{2} + 2 x + 4 = \left(x + 1 - \sqrt{3} i\right) \left(x + 1 + \sqrt{3} i\right)$