Simplifying rational expressions

Example 1: Simplify, $\dfrac{2u+4}{6u^2+12u}$

Solution: In the numerator, the greatest common factor is $2$; in the denominator, it is $6u$. Factoring these out,

$\dfrac{2u+4}{6u^2+12u}=\dfrac{2(u+2)}{6u(u+2)}$

Canceling the factors $2$ and $u+2$,

$\dfrac{2u+4}{6u^2+12u}=\dfrac{1}{3u}.$

Example 2: Simplify, $\dfrac{5x^2-5x-60}{x^2+8x+15}$.

Solution: First find the greatest common factor (GCF) of the numerator and of the denominator. The GCF of the numerator is $5$, and the denominator has no common factor. Factoring out the $5$,

$\dfrac{5x^2-5x-60}{x^2+8x+15}=\dfrac{5(x^2-x-12)}{x^2+8x+15}$

There are two quadratic trinomials, one in the numerator and the other in the denominator. Factoring them,

$x^2-x-12=(x+3)(x-4)$ and

$x^2+8x+15=(x+3)(x+5)$

Substituting these into the expression,

$\dfrac{5x^2-5x-60}{x^2+8x+15}=\dfrac{5(x+3)(x-4)}{(x+3)(x+5)}$

Canceling out $x+3$,

$\hphantom{00000}=\boxed{\dfrac{5(x-4)}{x+5}}$

Example 3: Simplify, $\dfrac{3y^2-48}{3y^2+7y-20}$

Solution:

In the numerator, the GCF is $3$. Factoring this out,

$\dfrac{3y^2-48}{3y^2+7y-20}=\dfrac{3(y^2-16)}{3y^2+7y-20}$

Factoring the binomial in the numerator and the trinomial in the denominator:

$\hphantom{00000}=\dfrac{3(y+4)(y-4)}{(3y-5)(y+4)}$

Canceling out $y+4$, we get

$\hphantom{00000}=\boxed{\dfrac{3(y-4)}{3y-5}}$