# How do you find all the zeros of f(x) = 9x^3 - 45x^2 + 36x?

Apr 9, 2016

$x = 0 , 1 , 4$

#### Explanation:

Factorise $f \left(x\right)$ by taking out $9 x$, the greatest common factor, and going from there,

$f \left(x\right) = 9 {x}^{3} - 45 {x}^{2} + 36 x$
$= 9 x \left({x}^{2} - 5 x + 4\right)$
$= 9 x \left(x - 4\right) \left(x - 1\right)$

One of these three parts has to equal $0$ for the whole thing to be $0$, so you'll end up with three answers.

$9 x = 0$
$x = \frac{0}{9} = 0$

$x - 4 = 0$
$x = 4$

$x - 1 = 0$
$x = 1$