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

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.