# Find the zeros of the function y=(x+2)^2 (x-5)^5 ?

Feb 14, 2018

$y = 0$ where $x = - 2$ and $x = 5$

#### Explanation:

We know that, if two numbers multiplied by each other equal zero, that one of the two numbers has to be zero. To find the zeros, set this equation equal to zero and solve for the $x$ values:

$0 = {\left(x + 2\right)}^{2} {\left(x - 5\right)}^{5}$ If this equation is true, we know that either
${\left(x + 2\right)}^{2} = 0$ or ${\left(x - 5\right)}^{5} = 0$.

Solve for $x$ in both cases:
${\left(x + 2\right)}^{2} = 0$
$\sqrt{{\left(x + 2\right)}^{2}} = \sqrt{0}$
$x + 2 = 0$
$x = - 2$

And
${\left(x - 5\right)}^{5} = 0$
$\sqrt[5]{x - 5} = \sqrt[5]{0}$
$x - 5 = 0$
$x = 5$

I hope that helps!