## Precalculus (6th Edition)

If $p=-4$, $q=8$ and $r=-10$, then by substituting in the given expression, we get: $\frac{5q+2(1+p)^3}{r+3}=\frac{5(8)+2(1+(-4))^3}{(-10)+3}=\frac{40+2(-3)^3}{-7}=\frac{40+2(-27)}{-7}=\frac{-14}{-7}=2$