# How do you find all zeros of the function f(x)=(x+5)^3(x-4)(x+1)^2?

Jul 6, 2018

color(blue)(x=-5,x=-5,x=-5 , x=4, x=-1

#### Explanation:

We first equate ${\left(x + 5\right)}^{3} \left(x - 4\right) \left(x + 1\right)$ to zero:

$\left(x + 5\right) \left(x + 5\right) \left(x + 5\right) \left(x - 4\right) \left(x + 1\right) = 0$

We just find values of x that make each bracket in turn zero.

Solutions here are:

$x = - 5 , x = - 5 , x = - 5 , x = 4 , x = - 1$

The graph confirms this: