Solve (-2-x)^2+3/-2-x+1+x^2+3/x+1=

$$\frac { ( - 2 - x ) ^ { 2 } + 3 } { - 2 - x + 1 } + \frac { x ^ { 2 } + 3 } { x + 1 } =$$

Evaluate

$$-4$$

Short Solution Steps

Add -2 and 1 to get -1.

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

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

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

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

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

Do the multiplications in -\left(\left(-2-x\right)^{2}+3\right)+x^{2}+3.

$$\frac{-4-4x-x^{2}-3+x^{2}+3}{x+1}$$

Combine like terms in -4-4x-x^{2}-3+x^{2}+3.

$$\frac{-4-4x}{x+1}$$

Factor the expressions that are not already factored in \frac{-4-4x}{x+1}.

$$\frac{4\left(-x-1\right)}{x+1}$$

Extract the negative sign in -1-x.

$$\frac{-4\left(x+1\right)}{x+1}$$

Cancel out x+1 in both numerator and denominator.

$$-4$$

Factor

$$-4$$