What are the zeros of the polynomial function f(x) = x^3 – x^2 - 12x?

Mar 15, 2016

The zeros occur at $x \in \left\{0 , 4 , - 3\right\}$

Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} f \left(x\right) = {x}^{3} - {x}^{2} - 12 x$

Factoring:
$\textcolor{w h i t e}{\text{XXX}} f \left(x\right) = x \left({x}^{2} - x - 12\right)$

$\textcolor{w h i t e}{\text{XXXXX}} = x \left(x - 4\right) \left(x + 3\right)$

The zeroes of a function are the values of $x$ for which the function's value is zero.

$\textcolor{w h i t e}{\text{XXX}} f \left(x\right) = x \left(x - 4\right) \left(x + 3\right) = 0$
implies
$\textcolor{w h i t e}{\text{XXX}}$either
$\textcolor{w h i t e}{\text{XXXXX}} x = 0$
$\textcolor{w h i t e}{\text{XXX}}$or
$\textcolor{w h i t e}{\text{XXXXX}} \left(x - 4\right) = 0 \rightarrow x = 4$
$\textcolor{w h i t e}{\text{XXX}}$or
$\textcolor{w h i t e}{\text{XXXXX}} \left(x + 3\right) = 0 \rightarrow x = - 3$