Solve -4n^2+n+2/2n^2+n+1/1-2

$$\frac { \frac { - 4 n ^ { 2 } + n + 2 } { 2 n ^ { 2 } + n + 1 } } { 1 - 2 }$$

Evaluate

$$-\frac{2+n-4n^{2}}{2n^{2}+n+1}$$

Short Solution Steps

Subtract 2 from 1 to get -1.

$$\frac{\frac{-4n^{2}+n+2}{2n^{2}+n+1}}{-1}$$

Express \frac{\frac{-4n^{2}+n+2}{2n^{2}+n+1}}{-1} as a single fraction.

$$\frac{-4n^{2}+n+2}{\left(2n^{2}+n+1\right)\left(-1\right)}$$

Factor the expressions that are not already factored.

$$\frac{-4\left(n-\left(-\frac{1}{8}\sqrt{33}+\frac{1}{8}\right)\right)\left(n-\left(\frac{1}{8}\sqrt{33}+\frac{1}{8}\right)\right)}{-\left(2n^{2}+n+1\right)}$$

Cancel out -1 in both numerator and denominator.

$$\frac{4\left(n-\left(-\frac{1}{8}\sqrt{33}+\frac{1}{8}\right)\right)\left(n-\left(\frac{1}{8}\sqrt{33}+\frac{1}{8}\right)\right)}{2n^{2}+n+1}$$

Expand the expression.

$$\frac{4n^{2}-n-2}{2n^{2}+n+1}$$

Expand

$$-\frac{2+n-4n^{2}}{2n^{2}+n+1}$$

Short Solution Steps

Subtract 2 from 1 to get -1.

$$\frac{\frac{-4n^{2}+n+2}{2n^{2}+n+1}}{-1}$$

Express \frac{\frac{-4n^{2}+n+2}{2n^{2}+n+1}}{-1} as a single fraction.

$$\frac{-4n^{2}+n+2}{\left(2n^{2}+n+1\right)\left(-1\right)}$$

Factor the expressions that are not already factored.

$$\frac{-4\left(n-\left(-\frac{1}{8}\sqrt{33}+\frac{1}{8}\right)\right)\left(n-\left(\frac{1}{8}\sqrt{33}+\frac{1}{8}\right)\right)}{-\left(2n^{2}+n+1\right)}$$

Cancel out -1 in both numerator and denominator.

$$\frac{4\left(n-\left(-\frac{1}{8}\sqrt{33}+\frac{1}{8}\right)\right)\left(n-\left(\frac{1}{8}\sqrt{33}+\frac{1}{8}\right)\right)}{2n^{2}+n+1}$$

Expand the expression.

$$\frac{4n^{2}-n-2}{2n^{2}+n+1}$$