How do you simplify #(4x^2+11x+6)/(x^2-x-6) *( x^2+x-12)/(x^2+8+16)#?

1 Answer
Jul 25, 2015

Answer:

Factor all the quadratics and cancel matching terms to find:

#(4x^2+11x+6)/(x^2-x-6)*(x^2+x-12)/(x^2+8x+16)=(4x+3)/(x+4)#

with exclusions #x!=-2#, #x!=3#

Explanation:

#(4x^2+11x+6)/(x^2-x-6)*(x^2+x-12)/(x^2+8x+16)#

#=((4x^2+11x+6)(x^2+x-12))/((x^2-x-6)(x^2+8x+16))#

#=(cancel((x+2))(4x+3)cancel((x+4))cancel((x-3)))/(cancel((x-3))cancel((x+2))cancel((x+4))(x+4))#

#=(4x+3)/(x+4)#

with exclusions #x != -2#, #x != 3#

The values #x=-2# and #x=3# are excluded because if substituted into the original expression, we get #0/0# which is undefined, but with the simplified expression, we get a well defined value. These are removable singularities of the original expression.

Therefore the simplified expression is not equivalent to the original expression for those values of #x#.

The value #x=-4# is not excluded from the equality of expressions, because it is equally a singularity of the original and simplified expressions, namely a simple pole of order #1#.