Question #c25c8

1 Answer
May 29, 2017

In it's lowest form, the expression equals #(3(x+3))/(2x+1)#, or #(3x+9)/(2x+1)#

Explanation:

First off, we should factor #(3x²-27)/(2x²-5x-3)#, and this produces:

#(3(x-3)(x+3))/((2x+1)(x-3))#

Now cancel out the term #(x-3)# from the numerator and denominator:

#(3cancel((x-3))(x+3))/((2x+1)cancel((x-3)))#

#(3(x+3))/(2x+1)#

From here you can distribute the #3# to form #3x+9# or leave it as that.