# How do you find all the real and complex roots of x^4 - 2x^3 - 4x^2 - 8x - 32 = 0?

Jan 28, 2016

Use the rational root theorem, then divide by the factors found to find the remaining roots.

$x = - 2$, $x = 4$, $x = \pm 2 i$

#### Explanation:

Let $f \left(x\right) = {x}^{4} - 2 {x}^{3} - 4 {x}^{2} - 8 x - 32$

By the rational root theorem, the only possible rational roots of $f \left(x\right) = 0$ are expressible in the form $\frac{p}{q}$ for some integers $p$ and $q$ where $p$ is a divisor of the constant term $- 32$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational roots are factors of $32$:

$\pm 1$, $\pm 2$, $\pm 4$, $\pm 8$, $\pm 16$, $\pm 32$

Trying each of these in turn, we find:

$f \left(- 2\right) = 16 + 16 - 16 + 16 - 32 = 0$

$f \left(4\right) = 256 - 128 - 64 - 32 - 32 = 0$

So $x = - 2$ and $x = 4$ are roots and $f \left(x\right)$ has corresponding factors $\left(x + 2\right)$ and $\left(x - 4\right)$

${x}^{4} - 2 {x}^{3} - 4 {x}^{2} - 8 x - 32$

$= \left(x + 2\right) \left({x}^{3} - 4 {x}^{2} + 4 x - 16\right)$

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

$= \left(x + 2\right) \left(x - 4\right) \left(x - 2 i\right) \left(x + 2 i\right)$