How do you simplify #(x^2+6x-27)div(3x^2+27x)/(x+5)#?

1 Answer
Sep 19, 2016

Answer:

If #x!=-9# and #x!=3# then

#(x^2+6x-27)div(3x^2+27x)/(x+5)=color(green)((x^2+2x-15)/(3x))#

Explanation:

Note that
#color(white)("XXX")x^2+6x-27=(x+9)(x-3)#
and that
#color(white)("XXX")3x^x+27=3x(x+9)#

and since #q div a/b rArr q * b/a#

#(x^2+6x-27) div (3x^2+27x)/(x+5)#

#color(white)("XXX")=(cancel(x+9))(x-3)xx(x+5)/(3x(cancel(x+9)))color(white)("XXX")#
#color(white)("XXX")#provided #x+9!=0 and 3x!=0#

#color(white)("XXX")=(x^2+2x-15)/(3x)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Note that if #x=-9# or #x=3#
then #3x^2+27x=0# and the division is undefined