# How do you find all the zeros of -5x^2(x-9)(x+4)^3?

Jun 30, 2018

$x = 0 , 9$ and $- 4$.

#### Explanation:

Let's set all terms equal to zero:

$\textcolor{b l u e}{- 5 {x}^{2} = 0}$

We need to take the square root of both sides to get

$- 5 x = 0$

$\textcolor{b l u e}{x = 0}$

$\textcolor{p u r p \le}{x - 9 = 0 \implies x = 9}$

${\left(x + 4\right)}^{3} = 0$

In this case, we need to take the cube root of both sides. We get

$\textcolor{red}{x + 4 = 0 \implies x = - 4}$

Therefore, the zeroes are at

$x = 0 , 9$ and $- 4$.

Hope this helps!