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

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Answer 1 · 2016-10-30T03:07:11

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

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

Factor

$y = (x + 5) (x + 3)$ $y=(x+5)(x+3)$

Set each factor to zero and solve.

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

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

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

They each have a multiplicity of $1$ $color(red)1$ because the exponent of each factor is $1$ $color(red)1$ as seen below.

$y = {(x + 5)}^{1} {(x + 3)}^{1}$ $y=(x+5)^color(red)1(x+3)^color(red)1$

How do you find all the zeros of y=x2+8x+15y=x^2+8x+15 with its multiplicities?