# How do you find all the zeros of h(x) = x^4 + 10x^3 + 26x^2 + 10x + 25?

Mar 20, 2016

Use the rational root theorem to help find the first zero, then factor by it and by grouping to find the other roots.

#### Explanation:

$h \left(x\right) = {x}^{4} + 10 {x}^{3} + 26 {x}^{2} + 10 x + 25$

By the rational root theorem, any rational roots of $h \left(x\right) = 0$ must be expressible in the form $\frac{p}{q}$ for integers $p$ and $q$ where $p$ is a factor of the constant term $25$ and $q$ a factor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1$, $\pm 5$, $\pm 25$

In addition, note that all of the coefficients of $h \left(x\right)$ are positive. So it can have no positive zeros. So that just leaves the possibilities:

$- 1$, $- 5$, $- 25$

We find:

$h \left(- 1\right) = 1 - 10 + 26 - 10 + 25 = 32$

$h \left(- 5\right) = 625 - 1250 + 650 - 50 + 25 = 0$

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

${x}^{4} + 10 {x}^{3} + 26 {x}^{2} + 10 x + 25$

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

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

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

So $x = - 5$ is a double zero and the other zeros are $x = \pm i$