# How do you find all zeros with multiplicities of f(x)=6x^4-5x^3-9x^2?

Dec 20, 2017

$x = 0$ (multiplicity $2$) and $x = \frac{5 \pm \sqrt{241}}{12}$ (multiplicity $1$ for both)

#### Explanation:

First we can factor out an ${x}^{2}$:
$6 {x}^{4} - 5 {x}^{3} - 9 {x}^{2} = {x}^{2} \left(6 {x}^{2} - 5 x - 9\right)$

From this, it's quite clear that $x = 0$ is a solution, and it will have multiplicity $2$ because the factor is ${x}^{2}$.

To find the remaining solutions, we can use the quadratic formula to solve:
$6 {x}^{2} - 5 x - 9 = 0$

$x = \frac{5 \pm \sqrt{241}}{12}$

These solutions only occur once, so they both have a multiplicity of $1$.