# How do you find all the zeros of f(x) = x^3 + 13x^2 + 57x + 85?

May 5, 2016

$x = 5$ or $x = - 4 \pm i$

#### Explanation:

$f \left(x\right) = {x}^{3} + 13 {x}^{2} + 57 x + 85$

By the rational roots theorem, any rational zeros of $f \left(x\right)$ must be expressible in the form $\frac{p}{q}$ for integers $p$, $q$ with $p$ a divisor of the constant term $85$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1$, $\pm 5$, $\pm 17$, $\pm 85$

In addition, since all of the coefficients of $f \left(x\right)$ are positive, it has no positive zeros, so that leaves:

$- 1$, $- 5$, $- 17$, $- 85$

Trying each of these in turn we find:

$f \left(- 5\right) = - 125 + 325 - 285 + 85 = 0$

So $x = - 5$ is a zero and $\left(x + 5\right)$ a factor:

${x}^{3} + 13 {x}^{2} + 57 x + 85 = \left(x + 5\right) \left({x}^{2} + 8 x + 17\right)$

We can find the remaining two zeros by completing the square:

$0 = {x}^{2} + 8 x + 17$

$= {\left(x + 4\right)}^{2} - 16 + 17$

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

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

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

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

So $x = - 4 \pm i$