Algebra 2 (1st Edition)

$$\frac{\left(x+4\right)\left(x+1\right)}{x^2-4x-21}$$
Dividing the two expressions and then cancelling out like terms and factors in the numerator and the denominator, we find: $$\frac{\left(x^2+9x+20\right)}{x^2-11x+28}\times \frac{x^2-3x-4}{\left(x^2+8x+15\right)}\\ \frac{\left(x+4\right)\left(x+5\right)\left(x+1\right)\left(x-4\right)}{\left(x-4\right)\left(x-7\right)\left(x+3\right)\left(x+5\right)}\\ \frac{\left(x+4\right)\left(x+1\right)}{x^2-4x-21}$$