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# How do you factor and solve x^2+8x+15=0?

Find roots that when added come to the middle term coefficient and when multiplied come to the last term and you'll find that $x = - 3$ and $x = - 5$

#### Explanation:

The key to factoring these kind of quadratic equations is to find 2 roots that when added up equal the coefficient of the middle term and when multiplied equal the last term.

If I'm having problems finding it, I often list out the factors of the last term and see what adds up to the middle term, like this:

$1 + 15 = 16$
$3 + 5 = 8$

(when things get complicated with negative signs and coefficients in front of the first term, this method can really help keep things organized).

Ok - we know the roots are 3 and 5:

$\left(x + 3\right) \left(x + 5\right) = 0$

The reason we set the left side equal to 0 is because it's easy to multiply by 0 in one term and get 0 as the overall answer. So what we do now is find out what it'll take to make x+3 = 0 and what it'll take x+5 = 0.

Let's do the x+3 first:

$x + 3 = 0$
$x = - 3$

Now the x+5 term:

$x + 5 = 0$
$x = - 5$

May 19, 2016

$\left(x + 3\right) \left(x + 5\right) = 0$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow \left(x = - 3\right) \mathmr{and} \left(x = - 5\right)$

#### Explanation:

Since $3 \times 5 = 15$ and $3 + 5 = 8$

${x}^{2} + 8 x + 15$ can be factored as $\left(x + 3\right) \left(x + 5\right)$

And since ${x}^{2} + 8 x + 15 = 0$
$\textcolor{w h i t e}{\text{XXX}} \Rightarrow \left(x + 3\right) \left(x + 5\right) = 0$

$\textcolor{w h i t e}{\text{XXX}} \Rightarrow$ either $\left(x + 3\right) = 0 \textcolor{w h i t e}{\text{XX")rarrcolor(white)("XX}} x = - 3$
$\textcolor{w h i t e}{\text{XXXXXx}}$or $\textcolor{w h i t e}{\text{XX")(x+5)=0color(white)("XX")rarrcolor(white)("XX}} x = - 5$

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