How do you simplify $(q^2+14q+48)/(q^2+q-30)$ ?

Question

1603 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-14T23:24:38

$(q+8)/(q-5)$ for $q != -6$ and $q != 5$

Notice that
$q^2+14q+48 = (q+6)(q+8)$
and
$q^2+q-30 = (q+6)(q-5)$

Therefore,
$(q^2+14q+48)/(q^2+q-30) = ((q+6)(q+8))/((q+6)(q-5))$

This expression is not defined for
$q = -6$ and $q = 5$

For all other values of $q$ we can reduce it by $q+6$ to
$(q+8)/(q-5)$

How do you simplify (q^2+14q+48)/(q^2+q-30)(q^2+14q+48)/(q^2+q-30)?