# How do you find all the zeros of h(x)=x^4+6x^3+10x^2+6x+9?

May 30, 2016

#### Answer:

$x = - 3$ (multiplicity 2)

$x = \pm i$

#### Explanation:

$h \left(x\right) = {x}^{4} + 6 {x}^{3} + 10 {x}^{2} + 6 x + 9$

By the rational root theorem, any rational zeros of $h \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $9$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1$, $\pm 3$, $\pm 9$

In addition, note that all of the coefficients of $h \left(x\right)$ are positive. So it has no positive zeros. So the only possible rational zeros are:

$- 1$, $- 3$, $- 9$

We find:

$h \left(- 1\right) = 1 - 6 + 10 - 6 + 9 = 8$

$h \left(- 3\right) = 81 - 162 + 90 - 18 + 9 = 0$

So $x = - 3$ is a zero and $\left(x + 3\right)$ a factor:

${x}^{4} + 6 {x}^{3} + 10 {x}^{2} + 6 x + 9 = \left(x + 3\right) \left({x}^{3} + 3 {x}^{2} + x + 3\right)$

The remaining cubic factors by grouping:

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

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

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

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

$= \left(x - i\right) \left(x + i\right) \left(x + 3\right)$

So $x = - 3$ is a zero again, giving it a total multiplicity $2$ and the other zeros are $x = \pm i$