## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$$\frac{1}{\left(z+1\right)\left(4z+3\right)}$$
Recall, one never divides by a fraction. Rather, if we want to divide by a fraction, we multiply by the reciprocal of the fraction. Recall, the reciprocal of a fraction is that fraction with the numerator and the denominator switched. Thus, we find: $$\frac{z^2-2z+1}{\left(z^2-1\right)\left(4z^2-z-3\right)}\\ \frac{\left(z-1\right)^2}{\left(z-1\right)\left(4z+3\right)\left(z^2-1\right)}\\ \frac{1}{\left(z+1\right)\left(4z+3\right)}$$