# How do you find all the zeros of f(x) = x^3-4x^2+9x-36?

May 23, 2018

The zeros are:

$x = 4 \text{ }$ and $\text{ } x = \pm 3 i$

#### Explanation:

Given:

$f \left(x\right) = {x}^{3} - 4 {x}^{2} + 9 x - 36$

Note that the ratio between the first and second terms is the same as that between the third and fourth terms. So this cubic will factor by grouping:

${x}^{3} - 4 {x}^{2} + 9 x - 36 = \left({x}^{3} - 4 {x}^{2}\right) + \left(9 x - 36\right)$

$\textcolor{w h i t e}{{x}^{3} - 4 {x}^{2} + 9 x - 36} = {x}^{2} \left(x - 4\right) + 9 \left(x - 4\right)$

$\textcolor{w h i t e}{{x}^{3} - 4 {x}^{2} + 9 x - 36} = \left({x}^{2} + 9\right) \left(x - 4\right)$

$\textcolor{w h i t e}{{x}^{3} - 4 {x}^{2} + 9 x - 36} = \left({x}^{2} + {3}^{2}\right) \left(x - 4\right)$

$\textcolor{w h i t e}{{x}^{3} - 4 {x}^{2} + 9 x - 36} = \left({x}^{2} - {\left(3 i\right)}^{2}\right) \left(x - 4\right)$

$\textcolor{w h i t e}{{x}^{3} - 4 {x}^{2} + 9 x - 36} = \left(x - 3 i\right) \left(x + 3 i\right) \left(x - 4\right)$

So the zeros are:

$x = 4 \text{ }$ and $\text{ } x = \pm 3 i$