Solve 6+8/x+2*4/x-2

$$6 + \frac { 8 } { x + 2 } \cdot \frac { 4 } { x - 2 }$$

Evaluate

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

Short Solution Steps

Multiply \frac{8}{x+2} times \frac{4}{x-2} by multiplying numerator times numerator and denominator times denominator.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 6 times \frac{\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}.

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

Since \frac{6\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)} and \frac{8\times 4}{\left(x+2\right)\left(x-2\right)} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 6\left(x+2\right)\left(x-2\right)+8\times 4.

$$\frac{6x^{2}-12x+12x-24+32}{\left(x+2\right)\left(x-2\right)}$$

Combine like terms in 6x^{2}-12x+12x-24+32.

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

Expand \left(x+2\right)\left(x-2\right).

$$\frac{6x^{2}+8}{x^{2}-4}$$

Factor

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