Solve 2x^2+4x/x^3-4x+x^2-4/x^2+3x+2/2-x

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

Evaluate

$$\frac{x^{2}-4}{x\left(x+3\right)}$$

Short Solution Steps

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

$$\frac{2x\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{x^{2}-4}{x^{2}+3x}+\frac{2}{2-x}$$

Cancel out x\left(x+2\right) in both numerator and denominator.

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

Factor x^{2}+3x.

$$\frac{2}{x-2}+\frac{x^{2}-4}{x\left(x+3\right)}+\frac{2}{2-x}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x-2 and x\left(x+3\right) is x\left(x-2\right)\left(x+3\right). Multiply \frac{2}{x-2} times \frac{x\left(x+3\right)}{x\left(x+3\right)}. Multiply \frac{x^{2}-4}{x\left(x+3\right)} times \frac{x-2}{x-2}.

$$\frac{2x\left(x+3\right)}{x\left(x-2\right)\left(x+3\right)}+\frac{\left(x^{2}-4\right)\left(x-2\right)}{x\left(x-2\right)\left(x+3\right)}+\frac{2}{2-x}$$

Since \frac{2x\left(x+3\right)}{x\left(x-2\right)\left(x+3\right)} and \frac{\left(x^{2}-4\right)\left(x-2\right)}{x\left(x-2\right)\left(x+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{2x\left(x+3\right)+\left(x^{2}-4\right)\left(x-2\right)}{x\left(x-2\right)\left(x+3\right)}+\frac{2}{2-x}$$

Do the multiplications in 2x\left(x+3\right)+\left(x^{2}-4\right)\left(x-2\right).

$$\frac{2x^{2}+6x+x^{3}-2x^{2}-4x+8}{x\left(x-2\right)\left(x+3\right)}+\frac{2}{2-x}$$

Combine like terms in 2x^{2}+6x+x^{3}-2x^{2}-4x+8.

$$\frac{2x+x^{3}+8}{x\left(x-2\right)\left(x+3\right)}+\frac{2}{2-x}$$

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

$$\frac{2x+x^{3}+8}{x\left(x-2\right)\left(x+3\right)}+\frac{2\left(-1\right)x\left(x+3\right)}{x\left(x-2\right)\left(x+3\right)}$$

Since \frac{2x+x^{3}+8}{x\left(x-2\right)\left(x+3\right)} and \frac{2\left(-1\right)x\left(x+3\right)}{x\left(x-2\right)\left(x+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{2x+x^{3}+8+2\left(-1\right)x\left(x+3\right)}{x\left(x-2\right)\left(x+3\right)}$$

Do the multiplications in 2x+x^{3}+8+2\left(-1\right)x\left(x+3\right).

$$\frac{2x+x^{3}+8-2x^{2}-6x}{x\left(x-2\right)\left(x+3\right)}$$

Combine like terms in 2x+x^{3}+8-2x^{2}-6x.

$$\frac{-4x+x^{3}+8-2x^{2}}{x\left(x-2\right)\left(x+3\right)}$$

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

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

Cancel out x-2 in both numerator and denominator.

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

Expand x\left(x+3\right).

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

Consider \left(x-2\right)\left(x+2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

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

Expand

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

Short Solution Steps

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

$$\frac{2x\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{x^{2}-4}{x^{2}+3x}+\frac{2}{2-x}$$

Cancel out x\left(x+2\right) in both numerator and denominator.

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

Factor x^{2}+3x.

$$\frac{2}{x-2}+\frac{x^{2}-4}{x\left(x+3\right)}+\frac{2}{2-x}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x-2 and x\left(x+3\right) is x\left(x-2\right)\left(x+3\right). Multiply \frac{2}{x-2} times \frac{x\left(x+3\right)}{x\left(x+3\right)}. Multiply \frac{x^{2}-4}{x\left(x+3\right)} times \frac{x-2}{x-2}.

$$\frac{2x\left(x+3\right)}{x\left(x-2\right)\left(x+3\right)}+\frac{\left(x^{2}-4\right)\left(x-2\right)}{x\left(x-2\right)\left(x+3\right)}+\frac{2}{2-x}$$

Since \frac{2x\left(x+3\right)}{x\left(x-2\right)\left(x+3\right)} and \frac{\left(x^{2}-4\right)\left(x-2\right)}{x\left(x-2\right)\left(x+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{2x\left(x+3\right)+\left(x^{2}-4\right)\left(x-2\right)}{x\left(x-2\right)\left(x+3\right)}+\frac{2}{2-x}$$

Do the multiplications in 2x\left(x+3\right)+\left(x^{2}-4\right)\left(x-2\right).

$$\frac{2x^{2}+6x+x^{3}-2x^{2}-4x+8}{x\left(x-2\right)\left(x+3\right)}+\frac{2}{2-x}$$

Combine like terms in 2x^{2}+6x+x^{3}-2x^{2}-4x+8.

$$\frac{2x+x^{3}+8}{x\left(x-2\right)\left(x+3\right)}+\frac{2}{2-x}$$

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

$$\frac{2x+x^{3}+8}{x\left(x-2\right)\left(x+3\right)}+\frac{2\left(-1\right)x\left(x+3\right)}{x\left(x-2\right)\left(x+3\right)}$$

Since \frac{2x+x^{3}+8}{x\left(x-2\right)\left(x+3\right)} and \frac{2\left(-1\right)x\left(x+3\right)}{x\left(x-2\right)\left(x+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{2x+x^{3}+8+2\left(-1\right)x\left(x+3\right)}{x\left(x-2\right)\left(x+3\right)}$$

Do the multiplications in 2x+x^{3}+8+2\left(-1\right)x\left(x+3\right).

$$\frac{2x+x^{3}+8-2x^{2}-6x}{x\left(x-2\right)\left(x+3\right)}$$

Combine like terms in 2x+x^{3}+8-2x^{2}-6x.

$$\frac{-4x+x^{3}+8-2x^{2}}{x\left(x-2\right)\left(x+3\right)}$$

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

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

Cancel out x-2 in both numerator and denominator.

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

Expand x\left(x+3\right).

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

Consider \left(x-2\right)\left(x+2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

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