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

Sep 18, 2016

The zeros (where$f \left(x\right) = 0$) are located at $x = 5$ and $x = - 4$.

Explanation:

To find the zeros, or x intercepts, set the equation equal to zero.
$f \left(x\right) = 0$ at the x intercepts.

$2 \left(x - 5\right) {\left(x + 4\right)}^{2} = 0$

$x - 5 = 0$ and $x + 4 = 0$

$x = 5$ and $x = - 4$

The zero at $x = 5$ has a multiplicity of one, because the exponent on the factor $\left(x - 5\right)$ is one. A zero with odd multiplicity indicates the graph crosses the x-axis at that point.

The zero at $x = - 4$ has a multiplicity of two, because the exponent on the factor ${\left(x - 4\right)}^{2}$ is two. A zero with even multiplicity indicates the graph just touches the x axis and then turns back in the same direction, i.e. the graph does not cross the x-axis.