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

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

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Answer 1 · 2015-12-09T12:26:08

$x = 0, 2$ $x=0,2$

Explanation:

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

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

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

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

We can say that any of these are true:

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

or

${(x - 2)}^{2} = 0$ $(x-2)^2=0$

Solve for both of these:

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

${(x - 2)}^{2} = 0$ $(x-2)^2=0$
$x - 2 = 0$ $x-2=0$
$x = 2$ $x=2$

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

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 $0$ $0$ and $2$ $2$ .

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Explanation:

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