# How do you find all zeros of f(x) = x^3 + 2x^2 - 5x - 6?

Jun 1, 2015

With no immediately obvious solutions, one way to attempt this is to plot the graph of the function.
Note that this does not guarantee that the x-intercepts will be exactly as they seem but it gives us some values to consider.
graph{x^3+2x^2-5x-6 [-10, 10, -5, 5]}
The graph of $f \left(x\right) = {x}^{3} + 2 x - 5 x - 6$
suggests that the solution set might be $x \epsilon \left\{- 3 , - 1 , + 2\right\}$

We can check this by using these values for a factoring:
$\textcolor{w h i t e}{\text{XXXXX}}$$\left(x + 3\right) \left(x + 1\right) \left(x - 2\right)$

Multiplying these possible factors we find that, in fact,
$\textcolor{w h i t e}{\text{XXXXX}}$$\left(x + 3\right) \left(x + 1\right) \left(x - 2\right) = {x}^{3} + 2 {x}^{2} - 5 x - 6$
so
the values previously considered to be merely "possible" have been verified as correct.