# How do you find all the zeros of f(x)=x^4+6x^3+14x^2+54x+45?

Aug 8, 2016

$f \left(x\right)$ has zeros $- 1 , - 5 , \pm 3 i$

#### Explanation:

$f \left(x\right) = {x}^{4} + 6 {x}^{3} + 14 {x}^{2} + 54 x + 45$

First note that:

$f \left(- 1\right) = 1 - 6 + 14 - 54 + 45 = 0$

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

${x}^{4} + 6 {x}^{3} + 14 {x}^{2} + 54 x + 45 = \left(x + 1\right) \left({x}^{3} + 5 {x}^{2} + 9 x + 45\right)$

Looking at the remaining cubic, note that the ratio between the first and second terms is the same as that between the third and fourth terms. So this cubic will factor by grouping...

${x}^{3} + 5 {x}^{2} + 9 x + 45$

$= \left({x}^{3} + 5 {x}^{2}\right) + \left(9 x + 45\right)$

$= {x}^{2} \left(x + 5\right) + 9 \left(x + 5\right)$

$= \left({x}^{2} + 9\right) \left(x + 5\right)$

Hence zeros $x = - 5$ and $x = \pm \sqrt{- 9} = \pm 3 i$