Solve (a/a^2-4-8/a^2+2a)a^2-2a/4-a

$$( \frac { a } { a ^ { 2 } - 4 } - \frac { 8 } { a ^ { 2 } + 2 a } ) \frac { a ^ { 2 } - 2 a } { 4 - a }$$

Evaluate

$$-\frac{a-4}{a+2}$$

Short Solution Steps

Factor a^{2}-4. Factor a^{2}+2a.

$$\left(\frac{a}{\left(a-2\right)\left(a+2\right)}-\frac{8}{a\left(a+2\right)}\right)\times \frac{a^{2}-2a}{4-a}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(a-2\right)\left(a+2\right) and a\left(a+2\right) is a\left(a-2\right)\left(a+2\right). Multiply \frac{a}{\left(a-2\right)\left(a+2\right)} times \frac{a}{a}. Multiply \frac{8}{a\left(a+2\right)} times \frac{a-2}{a-2}.

$$\left(\frac{aa}{a\left(a-2\right)\left(a+2\right)}-\frac{8\left(a-2\right)}{a\left(a-2\right)\left(a+2\right)}\right)\times \frac{a^{2}-2a}{4-a}$$

Since \frac{aa}{a\left(a-2\right)\left(a+2\right)} and \frac{8\left(a-2\right)}{a\left(a-2\right)\left(a+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{aa-8\left(a-2\right)}{a\left(a-2\right)\left(a+2\right)}\times \frac{a^{2}-2a}{4-a}$$

Do the multiplications in aa-8\left(a-2\right).

$$\frac{a^{2}-8a+16}{a\left(a-2\right)\left(a+2\right)}\times \frac{a^{2}-2a}{4-a}$$

Multiply \frac{a^{2}-8a+16}{a\left(a-2\right)\left(a+2\right)} times \frac{a^{2}-2a}{4-a} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(a^{2}-8a+16\right)\left(a^{2}-2a\right)}{a\left(a-2\right)\left(a+2\right)\left(4-a\right)}$$

Factor the expressions that are not already factored.

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

Cancel out a\left(a-2\right) in both numerator and denominator.

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

Expand the expression.

$$\frac{a^{2}-8a+16}{-a^{2}+2a+8}$$

Factor the expressions that are not already factored.

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

Cancel out a-4 in both numerator and denominator.

$$\frac{a-4}{-a-2}$$

Expand

$$-\frac{a-4}{a+2}$$

Short Solution Steps

Factor a^{2}-4. Factor a^{2}+2a.

$$\left(\frac{a}{\left(a-2\right)\left(a+2\right)}-\frac{8}{a\left(a+2\right)}\right)\times \frac{a^{2}-2a}{4-a}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(a-2\right)\left(a+2\right) and a\left(a+2\right) is a\left(a-2\right)\left(a+2\right). Multiply \frac{a}{\left(a-2\right)\left(a+2\right)} times \frac{a}{a}. Multiply \frac{8}{a\left(a+2\right)} times \frac{a-2}{a-2}.

$$\left(\frac{aa}{a\left(a-2\right)\left(a+2\right)}-\frac{8\left(a-2\right)}{a\left(a-2\right)\left(a+2\right)}\right)\times \frac{a^{2}-2a}{4-a}$$

Since \frac{aa}{a\left(a-2\right)\left(a+2\right)} and \frac{8\left(a-2\right)}{a\left(a-2\right)\left(a+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{aa-8\left(a-2\right)}{a\left(a-2\right)\left(a+2\right)}\times \frac{a^{2}-2a}{4-a}$$

Do the multiplications in aa-8\left(a-2\right).

$$\frac{a^{2}-8a+16}{a\left(a-2\right)\left(a+2\right)}\times \frac{a^{2}-2a}{4-a}$$

Multiply \frac{a^{2}-8a+16}{a\left(a-2\right)\left(a+2\right)} times \frac{a^{2}-2a}{4-a} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(a^{2}-8a+16\right)\left(a^{2}-2a\right)}{a\left(a-2\right)\left(a+2\right)\left(4-a\right)}$$

Factor the expressions that are not already factored.

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

Cancel out a\left(a-2\right) in both numerator and denominator.

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

Expand the expression.

$$\frac{a^{2}-8a+16}{-a^{2}+2a+8}$$

Factor the expressions that are not already factored.

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

Cancel out a-4 in both numerator and denominator.

$$\frac{a-4}{-a-2}$$