# How do you find zeros of y=5x^3-5x?

$x = - 1 , 0 , 1$

#### Explanation:

$y = 5 {x}^{3} - 5 x$

We can factor out a $5 x$ from both terms, so we get:

$y = 5 x \left({x}^{2} - 1\right)$

And now let's factor the expression within the brackets:

$y = 5 x \left(x + 1\right) \left(x - 1\right)$

We set it all equal to 0:

$0 = 5 x \left(x + 1\right) \left(x - 1\right)$

And we can now find the zeros:

$5 x = 0$

$x = 0$

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$x + 1 = 0$

$x = - 1$

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$x - 1 = 0$

$x = 1$

So $x = - 1 , 0 , 1$