How do you find all the zeros of y=x^2+8x+15 with its multiplicities?

Oct 30, 2016

The zeros are located at $x = - 5$ and $x = - 3$. They both have a multiplicity of $1$.

Explanation:

Find the zeros and multiplicities of the zeros of
$y = {x}^{2} + 8 x + 15$

Factor

$y = \left(x + 5\right) \left(x + 3\right)$

Set each factor to zero and solve.

$x + 5 = 0$ and $x + 3 = 0$

$x = - 5$ and $x = - 3$

The zeros are located at $x = - 5$ and $x = - 3$.

They each have a multiplicity of $\textcolor{red}{1}$ because the exponent of each factor is $\textcolor{red}{1}$ as seen below.

$y = {\left(x + 5\right)}^{\textcolor{red}{1}} {\left(x + 3\right)}^{\textcolor{red}{1}}$