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

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Answer 1 · 2017-11-08T21:21:24

$5/(x+7)$ ; $x!=-7$

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

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

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

$=5/(x+7)$

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 $x+7!=0$
$x!=-7$ so we exclude -7

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