Solve g(z)=z^3+4z/z^2-16divz^2+3z-18/z^2+2z-24

$$g ( z ) = \frac { z ^ { 3 } + 4 z } { z ^ { 2 } - 16 } \div \frac { z ^ { 2 } + 3 z - 18 } { z ^ { 2 } + 2 z - 24 }$$

Evaluate

$$\frac{z\left(z^{2}+4\right)}{\left(z-3\right)\left(z+4\right)}$$

Short Solution Steps

Divide \frac{z^{3}+4z}{z^{2}-16} by \frac{z^{2}+3z-18}{z^{2}+2z-24} by multiplying \frac{z^{3}+4z}{z^{2}-16} by the reciprocal of \frac{z^{2}+3z-18}{z^{2}+2z-24}.

$$\frac{\left(z^{3}+4z\right)\left(z^{2}+2z-24\right)}{\left(z^{2}-16\right)\left(z^{2}+3z-18\right)}$$

Factor the expressions that are not already factored.

$$\frac{z\left(z-4\right)\left(z+6\right)\left(z^{2}+4\right)}{\left(z-4\right)\left(z-3\right)\left(z+4\right)\left(z+6\right)}$$

Cancel out \left(z-4\right)\left(z+6\right) in both numerator and denominator.

$$\frac{z\left(z^{2}+4\right)}{\left(z-3\right)\left(z+4\right)}$$

Expand the expression.

$$\frac{z^{3}+4z}{z^{2}+z-12}$$

Expand

$$\frac{z^{3}+4z}{\left(z-3\right)\left(z+4\right)}$$

Short Solution Steps

Divide \frac{z^{3}+4z}{z^{2}-16} by \frac{z^{2}+3z-18}{z^{2}+2z-24} by multiplying \frac{z^{3}+4z}{z^{2}-16} by the reciprocal of \frac{z^{2}+3z-18}{z^{2}+2z-24}.

$$\frac{\left(z^{3}+4z\right)\left(z^{2}+2z-24\right)}{\left(z^{2}-16\right)\left(z^{2}+3z-18\right)}$$

Factor the expressions that are not already factored.

$$\frac{z\left(z-4\right)\left(z+6\right)\left(z^{2}+4\right)}{\left(z-4\right)\left(z-3\right)\left(z+4\right)\left(z+6\right)}$$

Cancel out \left(z-4\right)\left(z+6\right) in both numerator and denominator.

$$\frac{z\left(z^{2}+4\right)}{\left(z-3\right)\left(z+4\right)}$$

Expand the expression.

$$\frac{z^{3}+4z}{z^{2}+z-12}$$