## Precalculus: Mathematics for Calculus, 7th Edition

$\dfrac{2x+1}{2x^{2}+x-15}\div\dfrac{6x^{2}-x-2}{x+3}=\dfrac{1}{(2x-5)(3x-2)}$
$\dfrac{2x+1}{2x^{2}+x-15}\div\dfrac{6x^{2}-x-2}{x+3}$ Factor the denominator of the first fraction and the numerator of the second fraction: $\dfrac{2x+1}{(2x-5)(x+3)}\div\dfrac{(3x-2)(2x+1)}{x+3}=...$ Evaluate the division: $...=\dfrac{(2x+1)(x+3)}{(2x-5)(x+3)(3x-2)(2x+1)}=...$ Simplify: $...=\dfrac{1}{(2x-5)(3x-2)}$