# How do you find all zeros with multiplicities of f(x)=x^4-9x^2-4x+12?

Feb 25, 2017

$f \left(x\right)$ has zeros:

$1$ with multiplicity $1$

$3$ with multiplicity $1$

$- 2$ with multiplicity $2$

#### Explanation:

Given:

$f \left(x\right) = {x}^{4} - 9 {x}^{2} - 4 x + 12$

First note that the sum of the coefficients is $0$.

That is:

$1 - 9 - 4 + 12 = 0$

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

${x}^{4} - 9 {x}^{2} - 4 x + 12 = \left(x - 1\right) \left({x}^{3} + {x}^{2} - 8 x - 12\right)$

By the rational roots theorem, any rational zeros of ${x}^{3} + {x}^{2} - 8 x - 12$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 12$ and $q$ a divisor of the coefficient $1$ of the leading term.

So the only possible rational zeros of this cubic are:

$\pm 1 , \pm 2 , \pm 3 , \pm 4 , \pm 6 , \pm 12$

We find:

${\textcolor{b l u e}{3}}^{3} + {\textcolor{b l u e}{3}}^{2} - 8 \cdot \textcolor{b l u e}{3} - 12 = 27 + 9 - 24 - 12 = 0$

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

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

The remaining quadratic factor can be recognised as a perfect square trinomial. Like $144 = {12}^{2}$ we find:

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

Hence the other root is $- 2$ with multiplicity $2$.