Back to school - Finding the zeros/domain It's been a while since I've had to do math and I've been stuck around a problem for two good hours. I hate asking questions but I can't figure it out. 
I have the following problem: 
Find the zero/domain of 
$$ f(x) = \frac{9x^3-4x}{(x-3)(x^2-2x+1)} $$
So far, I've been able to find the domain by doing : 
$$ = \frac{9x^3-4x}{(x-3)(x-1)(x-1)} $$
$$ 0 = (x-1) $$
$$ 0 = (x-3) $$
But then I get stuck with the zero
$$ = \frac{x(9x^2-4)}{(x-3)(x-1)(x-1)} $$
That's where I'm stuck. What can I do with the $ x(9x^2-4) $
 A: To find the zero(s) of a function, we're after a solution to $f(x) = 0$. Here, the only possible solution to this is $$ x(9x^2-4) = 0$$
From this we can immediately see that one of the solutions is $ x = 0$. Factorising $9x^2 - 4$ gives $$(3x-2)(3x+2)$$
And therefore $$ (3x-2)(3x+2) = 0$$ Which gives another two solutions, $ x = \frac{2}{3}$ and $ x = -\frac{2}{3} $.
Therefore the solutions to $ f(x) = 0 $ are $ x = 0$ ,  $ x = \frac{2}{3}$, and $ x = -\frac{2}{3} $.
A: $$f(x)=\dfrac{x(9x^2-4)}{(x-3)(x-1)(x-1)}=0\,\,\text{ if the numerator is equal to $0.$}$$
So we just have to solve the equation $x(9x^2-4)=0$ which means that either $x=0$ or $9x^2-4=0$. The latter is a quadratic equation. And to solve it we use the quadratic formula which tells us that if $ax^2+bx+c=0$ where $a\neq0$ is a quadratic then its solutions are determined by the formula: $$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
We set $a=9$, $b=0$ and  $c=-4$ to get: $$x=\dfrac{\pm\sqrt{-4\cdot 9\cdot (-4)}}{2\cdot 9}=\dfrac{\pm\sqrt{144}}{18}=\pm \dfrac{12}{18}=\pm \dfrac{2}{3}.$$
Therefore the set of zeros of the function $f(x)$ is: $$\color{grey}{{\displaystyle\color{black}{\text{Set of zeros of $f(x)$}=\left\{0,\dfrac23,-\dfrac23\right\}}}}$$
I hope this helps.
Best wishes, $\mathcal H$akim.
