# How do you find all the zeros of f(x)= x^4+x^3+2x^2+4x-8 with its multiplicities?

Aug 3, 2016

$f \left(x\right)$ has zeros $1$, $- 2$ and $\pm 2 i$

#### Explanation:

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

First note that the sum of the coefficients of $f \left(x\right)$ is zero. That is:

$1 + 1 + 2 + 4 - 8 = 0$

Hence $f \left(1\right) = 0$, $x = 1$ is a zero of $f \left(x\right)$ and $\left(x - 1\right)$ is a factor:

${x}^{4} + {x}^{3} + 2 {x}^{2} + 4 x - 8 = \left(x - 1\right) \left({x}^{3} + 2 {x}^{2} + 4 x + 8\right)$

The remaining cubic is very regular and can be factored in a number of different ways. Let's factor it by grouping:

${x}^{3} + 2 {x}^{2} + 4 x + 8$

$= \left({x}^{3} + 2 {x}^{2}\right) + \left(4 x + 8\right)$

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

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

So $x = - 2$ is a zero.

The remaining quadratic has no Real zeros since ${x}^{2} + 4 \ge 4 > 0$ for any Real value of $x$, but it does have Complex zeros:

$x = \pm \sqrt{- 4} = \pm 2 i$