Appendix A - Review - A.6 Rational Expressions - A.6 Assess Your Understanding - Page A53: 15

$-(x+7), \quad x\ne2$

By factoring both denominator and numerator and canceling the common factors, we obtain: \begin{align*} \require{cancel} \dfrac{x^2+5x-14}{2-x}&=\dfrac{(x-2)(x+7)}{-(x-2)}\\ \\&=\dfrac{\cancel{(x-2)}(x+7)}{-\cancel{(x-2)}}\\ \\&=-(x+7), \quad x\ne2 \end{align*}