Solve 2n^2-2n-12/12-6n-6n^2/frac{n^2-6n+9{3n^2+6n-9}}

$$\frac { \frac { 2 n ^ { 2 } - 2 n - 12 } { 12 - 6 n - 6 n ^ { 2 } } } { \frac { n ^ { 2 } - 6 n + 9 } { 3 n ^ { 2 } + 6 n - 9 } }$$

Evaluate

$$-\frac{n+3}{n-3}$$

Short Solution Steps

Divide \frac{2n^{2}-2n-12}{12-6n-6n^{2}} by \frac{n^{2}-6n+9}{3n^{2}+6n-9} by multiplying \frac{2n^{2}-2n-12}{12-6n-6n^{2}} by the reciprocal of \frac{n^{2}-6n+9}{3n^{2}+6n-9}.

$$\frac{\left(2n^{2}-2n-12\right)\left(3n^{2}+6n-9\right)}{\left(12-6n-6n^{2}\right)\left(n^{2}-6n+9\right)}$$

Factor the expressions that are not already factored.

$$\frac{2\times 3\left(n-3\right)\left(n-1\right)\left(n+2\right)\left(n+3\right)}{6\left(n+2\right)\left(-n+1\right)\left(n-3\right)^{2}}$$

Extract the negative sign in -1+n.

$$\frac{-2\times 3\left(n-3\right)\left(n+2\right)\left(n+3\right)\left(-n+1\right)}{6\left(n+2\right)\left(-n+1\right)\left(n-3\right)^{2}}$$

Cancel out 2\times 3\left(n-3\right)\left(n+2\right)\left(-n+1\right) in both numerator and denominator.

$$\frac{-\left(n+3\right)}{n-3}$$

Expand the expression.

$$\frac{-n-3}{n-3}$$

Expand

$$-\frac{n+3}{n-3}$$

Short Solution Steps

Divide \frac{2n^{2}-2n-12}{12-6n-6n^{2}} by \frac{n^{2}-6n+9}{3n^{2}+6n-9} by multiplying \frac{2n^{2}-2n-12}{12-6n-6n^{2}} by the reciprocal of \frac{n^{2}-6n+9}{3n^{2}+6n-9}.

$$\frac{\left(2n^{2}-2n-12\right)\left(3n^{2}+6n-9\right)}{\left(12-6n-6n^{2}\right)\left(n^{2}-6n+9\right)}$$

Factor the expressions that are not already factored.

$$\frac{2\times 3\left(n-3\right)\left(n-1\right)\left(n+2\right)\left(n+3\right)}{6\left(n+2\right)\left(-n+1\right)\left(n-3\right)^{2}}$$

Extract the negative sign in -1+n.

$$\frac{-2\times 3\left(n-3\right)\left(n+2\right)\left(n+3\right)\left(-n+1\right)}{6\left(n+2\right)\left(-n+1\right)\left(n-3\right)^{2}}$$

Cancel out 2\times 3\left(n-3\right)\left(n+2\right)\left(-n+1\right) in both numerator and denominator.

$$\frac{-\left(n+3\right)}{n-3}$$

Expand the expression.

$$\frac{-n-3}{n-3}$$