# How do you find all the zeros of f(x)=x^3+5x^2+x+5?

Aug 14, 2016

$f \left(x\right)$ has zeros $- 5$ and $\pm i$

#### Explanation:

Since the ratio of the first and second terms is the same as that between the third and fourth terms, this cubic will factor by grouping.

So we find:

${x}^{3} + 5 {x}^{2} + x + 5$

$= {x}^{2} \left(x + 5\right) + 1 \left(x + 5\right)$

$= \left({x}^{2} + 1\right) \left(x + 5\right)$

$= \left({x}^{2} - {i}^{2}\right) \left(x + 5\right)$

$= \left(x - i\right) \left(x + i\right) \left(x + 5\right)$

Hence zeros: $\pm i$ and $- 5$