Solve {l}{-9/v^2+23v+22-0/v^2+18v+19}{-9}{-9}

$$\left. \begin{array} { l } { \frac { - 9 } { v ^ { 2 } + 23 v + 22 } - \frac { 0 } { v ^ { 2 } + 18 v + 19 } } \\ { - 9 } \\ { - 9 } \end{array} \right.$$

Evaluate

$$-\frac{9\left(2v^{2}+46v+45\right)}{\left(v+1\right)\left(v+22\right)}$$

Short Solution Steps

Zero divided by any non-zero term gives zero.

$$\frac{-9}{v^{2}+23v+22}+0-9-9$$

Subtract 9 from 0 to get -9.

$$\frac{-9}{v^{2}+23v+22}-9-9$$

Subtract 9 from -9 to get -18.

$$\frac{-9}{v^{2}+23v+22}-18$$

Factor v^{2}+23v+22.

$$\frac{-9}{\left(v+1\right)\left(v+22\right)}-18$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 18 times \frac{\left(v+1\right)\left(v+22\right)}{\left(v+1\right)\left(v+22\right)}.

$$\frac{-9}{\left(v+1\right)\left(v+22\right)}-\frac{18\left(v+1\right)\left(v+22\right)}{\left(v+1\right)\left(v+22\right)}$$

Since \frac{-9}{\left(v+1\right)\left(v+22\right)} and \frac{18\left(v+1\right)\left(v+22\right)}{\left(v+1\right)\left(v+22\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-9-18\left(v+1\right)\left(v+22\right)}{\left(v+1\right)\left(v+22\right)}$$

Do the multiplications in -9-18\left(v+1\right)\left(v+22\right).

$$\frac{-9-18v^{2}-396v-18v-396}{\left(v+1\right)\left(v+22\right)}$$

Combine like terms in -9-18v^{2}-396v-18v-396.

$$\frac{-405-18v^{2}-414v}{\left(v+1\right)\left(v+22\right)}$$

Expand \left(v+1\right)\left(v+22\right).

$$\frac{-405-18v^{2}-414v}{v^{2}+23v+22}$$

Factor

$$-\frac{18\left(v-\frac{-\sqrt{439}-23}{2}\right)\left(v-\frac{\sqrt{439}-23}{2}\right)}{\left(v+1\right)\left(v+22\right)}$$