# How do you find the zeroes for f(x)=x^3(x-2)^2?

Dec 9, 2015

$x = 0 , 2$

#### Explanation:

A zero is any point where $f \left(x\right) = 0$.

So, we have to find when ${x}^{3} {\left(x - 2\right)}^{2} = 0$.

Notice how there are terms being multiplied by one another, equaling $0$. The only way things can have a product of $0$ is if one of the things itself IS $0$.

So, to solve this, we can split apart ${x}^{3} {\left(x - 2\right)}^{2}$.

We can say that any of these are true:

${x}^{3} = 0$

or

${\left(x - 2\right)}^{2} = 0$

Solve for both of these:

${x}^{3} = 0$
$x = 0$

${\left(x - 2\right)}^{2} = 0$
$x - 2 = 0$
$x = 2$

Therefore $x = 0 , 2$ because both instances will give us an answer of $0$.

Look at a graph:
graph{x^3(x-2)^2 [-3.96, 7.14, -1.19, 4.357]}
The two zeros are indeed located at $0$ and $2$.