# How do you find zeros of f(x)=x^3-16x?

Nov 10, 2016

The zeros are ${x}_{1} = 0$, ${x}_{2} = - 4$ and ${x}_{3} = 4$. See explanation

#### Explanation:

To find the zeros we have to solve the equation:

## ${x}^{3} - 16 x = 0$

${x}^{3} - 16 x = 0$

$x \cdot \left({x}^{2} - 16\right) = 0$

$x \cdot \left(x - 4\right) \cdot \left(x + 4\right) = 0$

The product of expressions is zero if any of expressions is zero:

$x = 0 \vee x - 4 = 0 \vee x + 4 = 0$

$x = 0 \vee x = 4 \vee x = - 4$

Finaly we can write that the equation has 3 solutions:

${x}_{1} = - 4$, ${x}_{2} = 0$ and ${x}_{3} = 4$