Solve n(x)=x^2+2x+4/x^3-8-9-x^2/x^2+x-6

$$n ( x ) = \frac { x ^ { 2 } + 2 x + 4 } { x ^ { 3 } - 8 } - \frac { 9 - x ^ { 2 } } { x ^ { 2 } + x - 6 }$$

Evaluate

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Short Solution Steps

Factor the expressions that are not already factored in \frac{x^{2}+2x+4}{x^{3}-8}.

$$\frac{x^{2}+2x+4}{\left(x-2\right)\left(x^{2}+2x+4\right)}-\frac{9-x^{2}}{x^{2}+x-6}$$

Cancel out x^{2}+2x+4 in both numerator and denominator.

$$\frac{1}{x-2}-\frac{9-x^{2}}{x^{2}+x-6}$$

Factor the expressions that are not already factored in \frac{9-x^{2}}{x^{2}+x-6}.

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

Extract the negative sign in -3-x.

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

Cancel out x+3 in both numerator and denominator.

$$\frac{1}{x-2}-\frac{-\left(x-3\right)}{x-2}$$

Since \frac{1}{x-2} and \frac{-\left(x-3\right)}{x-2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1-\left(-\left(x-3\right)\right)}{x-2}$$

Do the multiplications in 1-\left(-\left(x-3\right)\right).

$$\frac{1+x-3}{x-2}$$

Combine like terms in 1+x-3.

$$\frac{-2+x}{x-2}$$

Cancel out x-2 in both numerator and denominator.

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Factor

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