# How do you find all zeros with multiplicities of f(x)=x^4+2x^3-12x^2-40x-32?

Feb 25, 2017

$f \left(x\right)$ has zeros $4$ (with multiplicity $1$) and $- 2$ with multiplicity $3$.

#### Explanation:

Given:

$f \left(x\right) = {x}^{4} + 2 {x}^{3} - 12 {x}^{2} - 40 x - 32$

By the rational roots theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divosor of the constant term $- 32$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 4 , \pm 8 , \pm 16 , \pm 32$

Note also that the signs of the coefficients are in the pattern $+ + - - -$. By Descartes' Rule of Signs, sinc this has one change, we can deduce that $f \left(x\right)$ has one positive real zero. So let us look for that one first.

$f \left(4\right) = {4}^{4} + 2 \left({4}^{3}\right) - 12 \left({4}^{2}\right) - 40 \left(4\right) - 32 = 256 + 128 - 192 - 160 - 32 = 0$

So $x = 4$ is a zero and $\left(x - 4\right)$ a factor:

${x}^{4} + 2 {x}^{3} - 12 {x}^{2} - 40 x - 32 = \left(x - 4\right) \left({x}^{3} + 6 {x}^{2} + 12 x + 8\right)$

Let us try substituting $- 2$ in the remaining cubic factor:

${\left(- 2\right)}^{3} + 6 {\left(- 2\right)}^{2} + 12 \left(- 2\right) + 8 = - 8 + 24 - 24 + 8 = 0$

So $x = - 2$ is a zero and $\left(x + 2\right)$ a factor:

${x}^{3} + 6 {x}^{2} + 12 x + 8 = \left(x + 2\right) \left({x}^{2} + 4 x + 4\right)$

The remaining quadratic factor is a perfect square trinomial, as we might spot by noticing $144 = {12}^{2}$ and similarly:

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

So $x = - 2$ is a zero twice more.

graph{x^4+2x^3-12x^2-40x-32 [-10, 10, -160, 80]}