Solve z^2-25/2z^2+9z-5*4z^2-1/3z-15

$$\frac { z ^ { 2 } - 25 } { 2 z ^ { 2 } + 9 z - 5 } \cdot \frac { 4 z ^ { 2 } - 1 } { 3 z - 15 }$$

Evaluate

$$\frac{2z+1}{3}$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{z^{2}-25}{2z^{2}+9z-5}.

$$\frac{\left(z-5\right)\left(z+5\right)}{\left(2z-1\right)\left(z+5\right)}\times \frac{4z^{2}-1}{3z-15}$$

Cancel out z+5 in both numerator and denominator.

$$\frac{z-5}{2z-1}\times \frac{4z^{2}-1}{3z-15}$$

Multiply \frac{z-5}{2z-1} times \frac{4z^{2}-1}{3z-15} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(z-5\right)\left(4z^{2}-1\right)}{\left(2z-1\right)\left(3z-15\right)}$$

Factor the expressions that are not already factored.

$$\frac{\left(z-5\right)\left(2z-1\right)\left(2z+1\right)}{3\left(z-5\right)\left(2z-1\right)}$$

Cancel out \left(z-5\right)\left(2z-1\right) in both numerator and denominator.

$$\frac{2z+1}{3}$$

Expand

$$\frac{2z+1}{3}$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{z^{2}-25}{2z^{2}+9z-5}.

$$\frac{\left(z-5\right)\left(z+5\right)}{\left(2z-1\right)\left(z+5\right)}\times \frac{4z^{2}-1}{3z-15}$$

Cancel out z+5 in both numerator and denominator.

$$\frac{z-5}{2z-1}\times \frac{4z^{2}-1}{3z-15}$$

Multiply \frac{z-5}{2z-1} times \frac{4z^{2}-1}{3z-15} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(z-5\right)\left(4z^{2}-1\right)}{\left(2z-1\right)\left(3z-15\right)}$$

Factor the expressions that are not already factored.

$$\frac{\left(z-5\right)\left(2z-1\right)\left(2z+1\right)}{3\left(z-5\right)\left(2z-1\right)}$$

Cancel out \left(z-5\right)\left(2z-1\right) in both numerator and denominator.

$$\frac{2z+1}{3}$$