Solve -1(2x^2+4x+8)/(x^2+2x+4)*(x+2)(x-2)/-1(x-1)

$$\frac { - 1 ( 2 x ^ { 2 } + 4 x + 8 ) } { ( x ^ { 2 } + 2 x + 4 ) } \cdot \frac { ( x + 2 ) ( x - 2 ) } { - 1 ( x - 1 ) }$$

Evaluate

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

Short Solution Steps

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

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

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

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

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.

$$-2\times \frac{x^{2}-4}{-\left(x-1\right)}$$

To find the opposite of x-1, find the opposite of each term.

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

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

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

Use the distributive property to multiply -2 by x^{2}-4.

$$\frac{-2x^{2}+8}{-x+1}$$

Expand

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

Short Solution Steps

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

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

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

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

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.

$$-2\times \frac{x^{2}-4}{-\left(x-1\right)}$$

To find the opposite of x-1, find the opposite of each term.

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

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

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

Use the distributive property to multiply -2 by x^{2}-4.

$$\frac{-2x^{2}+8}{-x+1}$$