# How do you simplify (5x+15)/(x^2+10x+21) and then find the excluded values?

Nov 8, 2017

$\frac{5}{x + 7}$; $x \ne - 7$

#### Explanation:

Always. ALWAYS factorise an algebraic fraction (if you can). We can factorise the top to

$5 \left(x + 3\right)$.
This helps us out, we can be pretty sure the quadratic will have $\left(x + 3\right)$ as a factor. Sure enough, we get:

(5x+15)/(x^2+10x+21)=[5cancel((x+3))]/((cancel(x+3))(x+7)

$= \frac{5}{x + 7}$

In maths, dividing by zero does loads of crazy things, so we need to exclude the x that would make then denominator equal to zero.

let $x + 7 \ne 0$
$x \ne - 7$ so we exclude -7