# How do you find all zeros of the function f(x) = x^3 - 6x^2 + 9x?

Mar 2, 2016

$f \left(x\right) = 0 \text{ at "x=0" ; } x = + 3$

#### Explanation:

Notice that $x$ is in each term so we can do this:

$f \left(x\right) = x \left({x}^{2} - 6 x + 9\right)$

Notice that $3 \times 3 = 9 \text{ and that } 3 + 3 = 6$

But we have $- 6$ so it has to be $- 3 - 3 = - 6$

$f \left(x\right) = x {\left(x - 3\right)}^{2}$

so $f \left(x\right) = 0 \text{ at "x=0" ; } x = + 3$