Solve z^2-4/4z+8div2z^2-8z+8/8z+16

$$\frac { z ^ { 2 } - 4 } { 4 z + 8 } \div \frac { 2 z ^ { 2 } - 8 z + 8 } { 8 z + 16 }$$

Evaluate

$$\frac{z+2}{z-2}$$

Short Solution Steps

Divide \frac{z^{2}-4}{4z+8} by \frac{2z^{2}-8z+8}{8z+16} by multiplying \frac{z^{2}-4}{4z+8} by the reciprocal of \frac{2z^{2}-8z+8}{8z+16}.

$$\frac{\left(z^{2}-4\right)\left(8z+16\right)}{\left(4z+8\right)\left(2z^{2}-8z+8\right)}$$

Factor the expressions that are not already factored.

$$\frac{8\left(z-2\right)\left(z+2\right)^{2}}{2\times 4\left(z+2\right)\left(z-2\right)^{2}}$$

Cancel out 2\times 4\left(z-2\right)\left(z+2\right) in both numerator and denominator.

$$\frac{z+2}{z-2}$$

Expand

$$\frac{z+2}{z-2}$$

Short Solution Steps

Divide \frac{z^{2}-4}{4z+8} by \frac{2z^{2}-8z+8}{8z+16} by multiplying \frac{z^{2}-4}{4z+8} by the reciprocal of \frac{2z^{2}-8z+8}{8z+16}.

$$\frac{\left(z^{2}-4\right)\left(8z+16\right)}{\left(4z+8\right)\left(2z^{2}-8z+8\right)}$$

Factor the expressions that are not already factored.

$$\frac{8\left(z-2\right)\left(z+2\right)^{2}}{2\times 4\left(z+2\right)\left(z-2\right)^{2}}$$

Cancel out 2\times 4\left(z-2\right)\left(z+2\right) in both numerator and denominator.

$$\frac{z+2}{z-2}$$