# How do you find all zeros of the function f(x)=x^3+3x^2-34x+48?

Jul 15, 2016

The zeros are $2 , 3 ,$ and $- 8$.

#### Explanation:

A simple way to find the zeros of this function is to factor the polynomial. There are several ways to find these factors, but for this question, let's take a computer-based short-cut! We can look for the zeros by plotting the function, and if we are lucky we'll see the integer values that correspond to the zeros:

graph{x^3+3x^2-34x+48 [-10, 10, -20, 20]}

from this we can guess that the zeros are $2 , 3 ,$ and $- 8$ which makes our function:

$f \left(x\right) = \left(x - 2\right) \left(x - 3\right) \left(x + 8\right)$

if we multiply this out and regain our original cubic polynomial then our guesses are correct:

$f \left(x\right) = \left({x}^{2} - 5 x + 6\right) \left(x + 8\right) = {x}^{3} + 3 {x}^{2} - 34 x + 48$

therefore, our guesses are correct!