Solve 9/2z^2+6z-36-6z+2/2z^2+4z-30

$$\frac { 9 } { 2 z ^ { 2 } + 6 z - 36 } - \frac { 6 z + 2 } { 2 z ^ { 2 } + 4 z - 30 }$$

Evaluate

$$\frac{33-29z-6z^{2}}{2\left(z-3\right)\left(z+5\right)\left(z+6\right)}$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{6z+2}{2z^{2}+4z-30}.

$$\frac{9}{2z^{2}+6z-36}-\frac{2\left(3z+1\right)}{2\left(z-3\right)\left(z+5\right)}$$

Cancel out 2 in both numerator and denominator.

$$\frac{9}{2z^{2}+6z-36}-\frac{3z+1}{\left(z-3\right)\left(z+5\right)}$$

Factor 2z^{2}+6z-36.

$$\frac{9}{2\left(z-3\right)\left(z+6\right)}-\frac{3z+1}{\left(z-3\right)\left(z+5\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2\left(z-3\right)\left(z+6\right) and \left(z-3\right)\left(z+5\right) is 2\left(z-3\right)\left(z+5\right)\left(z+6\right). Multiply \frac{9}{2\left(z-3\right)\left(z+6\right)} times \frac{z+5}{z+5}. Multiply \frac{3z+1}{\left(z-3\right)\left(z+5\right)} times \frac{2\left(z+6\right)}{2\left(z+6\right)}.

$$\frac{9\left(z+5\right)}{2\left(z-3\right)\left(z+5\right)\left(z+6\right)}-\frac{\left(3z+1\right)\times 2\left(z+6\right)}{2\left(z-3\right)\left(z+5\right)\left(z+6\right)}$$

Since \frac{9\left(z+5\right)}{2\left(z-3\right)\left(z+5\right)\left(z+6\right)} and \frac{\left(3z+1\right)\times 2\left(z+6\right)}{2\left(z-3\right)\left(z+5\right)\left(z+6\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{9\left(z+5\right)-\left(3z+1\right)\times 2\left(z+6\right)}{2\left(z-3\right)\left(z+5\right)\left(z+6\right)}$$

Do the multiplications in 9\left(z+5\right)-\left(3z+1\right)\times 2\left(z+6\right).

$$\frac{9z+45-6z^{2}-36z-2z-12}{2\left(z-3\right)\left(z+5\right)\left(z+6\right)}$$

Combine like terms in 9z+45-6z^{2}-36z-2z-12.

$$\frac{-29z+33-6z^{2}}{2\left(z-3\right)\left(z+5\right)\left(z+6\right)}$$

Factor the expressions that are not already factored in \frac{-29z+33-6z^{2}}{2\left(z-3\right)\left(z+5\right)\left(z+6\right)}.

$$\frac{-6\left(z-\left(-\frac{1}{12}\sqrt{1633}-\frac{29}{12}\right)\right)\left(z-\left(\frac{1}{12}\sqrt{1633}-\frac{29}{12}\right)\right)}{2\left(z-3\right)\left(z+5\right)\left(z+6\right)}$$

Cancel out 2 in both numerator and denominator.

$$\frac{-3\left(z-\left(-\frac{1}{12}\sqrt{1633}-\frac{29}{12}\right)\right)\left(z-\left(\frac{1}{12}\sqrt{1633}-\frac{29}{12}\right)\right)}{\left(z-3\right)\left(z+5\right)\left(z+6\right)}$$

Expand \left(z-3\right)\left(z+5\right)\left(z+6\right).

$$\frac{-3\left(z-\left(-\frac{1}{12}\sqrt{1633}-\frac{29}{12}\right)\right)\left(z-\left(\frac{1}{12}\sqrt{1633}-\frac{29}{12}\right)\right)}{z^{3}+8z^{2}-3z-90}$$

To find the opposite of -\frac{1}{12}\sqrt{1633}-\frac{29}{12}, find the opposite of each term.

$$\frac{-3\left(z+\frac{1}{12}\sqrt{1633}+\frac{29}{12}\right)\left(z-\left(\frac{1}{12}\sqrt{1633}-\frac{29}{12}\right)\right)}{z^{3}+8z^{2}-3z-90}$$

To find the opposite of \frac{1}{12}\sqrt{1633}-\frac{29}{12}, find the opposite of each term.

$$\frac{-3\left(z+\frac{1}{12}\sqrt{1633}+\frac{29}{12}\right)\left(z-\frac{1}{12}\sqrt{1633}+\frac{29}{12}\right)}{z^{3}+8z^{2}-3z-90}$$

Use the distributive property to multiply -3 by z+\frac{1}{12}\sqrt{1633}+\frac{29}{12}.

$$\frac{\left(-3z-\frac{1}{4}\sqrt{1633}-\frac{29}{4}\right)\left(z-\frac{1}{12}\sqrt{1633}+\frac{29}{12}\right)}{z^{3}+8z^{2}-3z-90}$$

Use the distributive property to multiply -3z-\frac{1}{4}\sqrt{1633}-\frac{29}{4} by z-\frac{1}{12}\sqrt{1633}+\frac{29}{12} and combine like terms.

$$\frac{-3z^{2}-\frac{29}{2}z+\frac{1}{48}\left(\sqrt{1633}\right)^{2}-\frac{841}{48}}{z^{3}+8z^{2}-3z-90}$$

The square of \sqrt{1633} is 1633.

$$\frac{-3z^{2}-\frac{29}{2}z+\frac{1}{48}\times 1633-\frac{841}{48}}{z^{3}+8z^{2}-3z-90}$$

Multiply \frac{1}{48} and 1633 to get \frac{1633}{48}.

$$\frac{-3z^{2}-\frac{29}{2}z+\frac{1633}{48}-\frac{841}{48}}{z^{3}+8z^{2}-3z-90}$$

Subtract \frac{841}{48} from \frac{1633}{48} to get \frac{33}{2}.

$$\frac{-3z^{2}-\frac{29}{2}z+\frac{33}{2}}{z^{3}+8z^{2}-3z-90}$$

Expand

$$-\frac{6z^{2}+29z-33}{2\left(z-3\right)\left(z+5\right)\left(z+6\right)}$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{6z+2}{2z^{2}+4z-30}.

$$\frac{9}{2z^{2}+6z-36}-\frac{2\left(3z+1\right)}{2\left(z-3\right)\left(z+5\right)}$$

Cancel out 2 in both numerator and denominator.

$$\frac{9}{2z^{2}+6z-36}-\frac{3z+1}{\left(z-3\right)\left(z+5\right)}$$

Factor 2z^{2}+6z-36.

$$\frac{9}{2\left(z-3\right)\left(z+6\right)}-\frac{3z+1}{\left(z-3\right)\left(z+5\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2\left(z-3\right)\left(z+6\right) and \left(z-3\right)\left(z+5\right) is 2\left(z-3\right)\left(z+5\right)\left(z+6\right). Multiply \frac{9}{2\left(z-3\right)\left(z+6\right)} times \frac{z+5}{z+5}. Multiply \frac{3z+1}{\left(z-3\right)\left(z+5\right)} times \frac{2\left(z+6\right)}{2\left(z+6\right)}.

$$\frac{9\left(z+5\right)}{2\left(z-3\right)\left(z+5\right)\left(z+6\right)}-\frac{\left(3z+1\right)\times 2\left(z+6\right)}{2\left(z-3\right)\left(z+5\right)\left(z+6\right)}$$

Since \frac{9\left(z+5\right)}{2\left(z-3\right)\left(z+5\right)\left(z+6\right)} and \frac{\left(3z+1\right)\times 2\left(z+6\right)}{2\left(z-3\right)\left(z+5\right)\left(z+6\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{9\left(z+5\right)-\left(3z+1\right)\times 2\left(z+6\right)}{2\left(z-3\right)\left(z+5\right)\left(z+6\right)}$$

Do the multiplications in 9\left(z+5\right)-\left(3z+1\right)\times 2\left(z+6\right).

$$\frac{9z+45-6z^{2}-36z-2z-12}{2\left(z-3\right)\left(z+5\right)\left(z+6\right)}$$

Combine like terms in 9z+45-6z^{2}-36z-2z-12.

$$\frac{-29z+33-6z^{2}}{2\left(z-3\right)\left(z+5\right)\left(z+6\right)}$$

Factor the expressions that are not already factored in \frac{-29z+33-6z^{2}}{2\left(z-3\right)\left(z+5\right)\left(z+6\right)}.

$$\frac{-6\left(z-\left(-\frac{1}{12}\sqrt{1633}-\frac{29}{12}\right)\right)\left(z-\left(\frac{1}{12}\sqrt{1633}-\frac{29}{12}\right)\right)}{2\left(z-3\right)\left(z+5\right)\left(z+6\right)}$$

Cancel out 2 in both numerator and denominator.

$$\frac{-3\left(z-\left(-\frac{1}{12}\sqrt{1633}-\frac{29}{12}\right)\right)\left(z-\left(\frac{1}{12}\sqrt{1633}-\frac{29}{12}\right)\right)}{\left(z-3\right)\left(z+5\right)\left(z+6\right)}$$

Expand \left(z-3\right)\left(z+5\right)\left(z+6\right).

$$\frac{-3\left(z-\left(-\frac{1}{12}\sqrt{1633}-\frac{29}{12}\right)\right)\left(z-\left(\frac{1}{12}\sqrt{1633}-\frac{29}{12}\right)\right)}{z^{3}+8z^{2}-3z-90}$$

To find the opposite of -\frac{1}{12}\sqrt{1633}-\frac{29}{12}, find the opposite of each term.

$$\frac{-3\left(z+\frac{1}{12}\sqrt{1633}+\frac{29}{12}\right)\left(z-\left(\frac{1}{12}\sqrt{1633}-\frac{29}{12}\right)\right)}{z^{3}+8z^{2}-3z-90}$$

To find the opposite of \frac{1}{12}\sqrt{1633}-\frac{29}{12}, find the opposite of each term.

$$\frac{-3\left(z+\frac{1}{12}\sqrt{1633}+\frac{29}{12}\right)\left(z-\frac{1}{12}\sqrt{1633}+\frac{29}{12}\right)}{z^{3}+8z^{2}-3z-90}$$

Use the distributive property to multiply -3 by z+\frac{1}{12}\sqrt{1633}+\frac{29}{12}.

$$\frac{\left(-3z-\frac{1}{4}\sqrt{1633}-\frac{29}{4}\right)\left(z-\frac{1}{12}\sqrt{1633}+\frac{29}{12}\right)}{z^{3}+8z^{2}-3z-90}$$

Use the distributive property to multiply -3z-\frac{1}{4}\sqrt{1633}-\frac{29}{4} by z-\frac{1}{12}\sqrt{1633}+\frac{29}{12} and combine like terms.

$$\frac{-3z^{2}-\frac{29}{2}z+\frac{1}{48}\left(\sqrt{1633}\right)^{2}-\frac{841}{48}}{z^{3}+8z^{2}-3z-90}$$

The square of \sqrt{1633} is 1633.

$$\frac{-3z^{2}-\frac{29}{2}z+\frac{1}{48}\times 1633-\frac{841}{48}}{z^{3}+8z^{2}-3z-90}$$

Multiply \frac{1}{48} and 1633 to get \frac{1633}{48}.

$$\frac{-3z^{2}-\frac{29}{2}z+\frac{1633}{48}-\frac{841}{48}}{z^{3}+8z^{2}-3z-90}$$

Subtract \frac{841}{48} from \frac{1633}{48} to get \frac{33}{2}.

$$\frac{-3z^{2}-\frac{29}{2}z+\frac{33}{2}}{z^{3}+8z^{2}-3z-90}$$