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

1 Answer
Nov 8, 2017

Answer:

#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