## College Algebra (10th Edition)

$\displaystyle \frac{2(2x^{2}+5x-2)}{(x-2)(x+2)(x+3)}$
Step 1: Factor each denominator $x^{2}-4=\qquad$ ... a difference of squares, $=(x-2)(x+2)$ For $x^{2}+bx+c$, we search for factors of c whose sum is b: $x^{2}+x-6=\qquad$ ... +3 and -2 $=(x+3)(x-2)$ Step 2: The LCM is the product of each of these factors raised to a power equal to the greatest number of times that the factor occurs in the polynomials. LCM = $(x-2)(x+2)(x+3)$ Step 3: Write each rational expression using the LCM as the denominator. Simplify. $\displaystyle \frac{4x}{(x-2)(x+2)}\cdot\frac{(x+3)}{(x+3)}-\frac{2}{(x+3)(x-2)}\frac{(x+2)}{(x+2)}$ $=\displaystyle \frac{4x^{2}+12x-(2x+4)}{(x-2)(x+2)(x+3)}$ $=\displaystyle \frac{4x^{2}+10x-4}{(x-2)(x+2)(x+3)}$ $=\displaystyle \frac{2(2x^{2}+5x-2)}{(x-2)(x+2)(x+3)}$ For $ax^{2}+bx+c$ in the numerator, we search for factors of ac whose sum is b, and rewrite the $bx$ term: ... no two factors of -4 add to +5, so we leave it as it is.