How to find all zeros of f ( x ) = 3 x 3 -...

We must first determine the derivative of a function before we can determine its zeros. This one is simple; we'll just use the power rule to evaluate each phrase individually.
${\displaystyle f\left(x\right)=3x^3-12x^2+3x}$
${\displaystyle f\prime \left(x\right)=9x^2-24x+3}$
Then we set the derivative equal to zero and solve for x
${\displaystyle 9x^2-24x+3=0}$
Given that this function cannot be factored, we can enter this into our calculator to determine the two x values. Obtaining two values
${\displaystyle x=2.535}$ ${\displaystyle x=.131}$

First, factor out an x:
${\displaystyle f\left(x\right)=x(3x^2-12x+3x)}$
Already, you can figure out that when x zero, f(x) is zero, so ${\displaystyle x=0}$ is one solution.
Then, attempt to factor the quadratic. Unfortunately, in this case, you can't use factoring, so plug the numbers into the quadratic formula, which is ${\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2}a}$ in case you don't have it memorized:
${\displaystyle x=\frac{-(-12)\pm \sqrt{{(-12)}^2-4\left(3\right)\left(3\right)}}{2\left(3\right)}}$
${\displaystyle x=\frac{12\pm \sqrt{144-36}}{2\left(3\right)}}$
${\displaystyle x=\frac{12\pm \sqrt{108}}{6}}$
(The square root can be simplified to ${\displaystyle 6\sqrt{3}}$, because ${\displaystyle \sqrt{108}=\sqrt{6\cdot 6\cdot 3}={\sqrt{6}}^2\cdot \sqrt{3}=6\cdot \sqrt{3}}$)
${\displaystyle x=\frac{12\pm 6\sqrt{3}}{6}}$
A six can also be cancelled out:
${\displaystyle x=(2\pm \sqrt{3})}$
That means the zeroes are ${\displaystyle x=0,x=2+\sqrt{3},\text{and}x=2-\sqrt{3}}$