Question

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

Answer 1

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

Explanation:

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