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

Aug 11, 2018

$x = 0 \text{ }$ or $\text{ } x = \frac{5}{18} \pm \frac{\sqrt{61}}{18}$

all with multiplicity $1$

#### Explanation:

$0 = 36 f \left(x\right)$

$\textcolor{w h i t e}{0} = 36 \left(9 {x}^{3} - 5 {x}^{2} - x\right)$

$\textcolor{w h i t e}{0} = x \left(324 {x}^{2} - 180 x - 36\right)$

$\textcolor{w h i t e}{0} = x \left({\left(18 x\right)}^{2} - 2 \left(18 x\right) \left(5\right) + 25 - 61\right)$

$\textcolor{w h i t e}{0} = x \left({\left(18 x - 5\right)}^{2} - {\left(\sqrt{61}\right)}^{2}\right)$

$\textcolor{w h i t e}{0} = x \left(18 x - 5 - \sqrt{61}\right) \left(18 x - 5 + \sqrt{61}\right)$

Hence:

$x = 0$

or:

$18 x = 5 \pm \sqrt{61} \text{ }$ so $\text{ } x = \frac{5}{18} \pm \frac{\sqrt{61}}{18}$