# How do you find all zeros of f(x)=3x^3-12x^2+3x?

Mar 20, 2017

$x = 2.535$

$x = .131$

#### Explanation:

To find the zeros of a function, we must first find the derivative of the function. This one is not very difficult, we will just apply the power rule to each individual term

$f \left(x\right) = 3 {x}^{3} - 12 {x}^{2} + 3 x$
$f ' \left(x\right) = 9 {x}^{2} - 24 x + 3$

Then we set the derivative equal to zero and solve for x

$9 {x}^{2} - 24 x + 3 = 0$

We can plug this into our calculator to find the two x values seeing as how this function is not factorable. We get two values

$x = 2.535$ $x = .131$

Mar 20, 2017

$x = 0 , x = 2 + \sqrt{3} , \mathmr{and} x = 2 - \sqrt{3}$

#### Explanation:

First, factor out an x:
$f \left(x\right) = x \left(3 {x}^{2} - 12 x + 3 x\right)$
Already, you can figure out that when x zero, f(x) is zero, so $x = 0$ 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 $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2} a$ in case you don't have it memorized:
$x = \frac{- \left(- 12\right) \pm \sqrt{{\left(- 12\right)}^{2} - 4 \left(3\right) \left(3\right)}}{2 \left(3\right)}$
$x = \frac{12 \pm \sqrt{144 - 36}}{2 \left(3\right)}$
$x = \frac{12 \pm \sqrt{108}}{6}$
(The square root can be simplified to $6 \sqrt{3}$, because $\sqrt{108} = \sqrt{6 \cdot 6 \cdot 3} = {\sqrt{6}}^{2} \cdot \sqrt{3} = 6 \cdot \sqrt{3}$)
$x = \frac{12 \pm 6 \sqrt{3}}{6}$
A six can also be cancelled out:
$x = \left(2 \pm \sqrt{3}\right)$
That means the zeroes are $x = 0 , x = 2 + \sqrt{3} , \mathmr{and} x = 2 - \sqrt{3}$.