Solve frac{4a^{2}+2a}{a^{3}-12a-16}-2a^2-4/a^3+4a^2+4a-1/a-4a-4/a^2-2a-8

$$\frac { 4 a ^ { 2 } + 2 a } { a ^ { 3 } - 12 a - 16 } - \frac { 2 a ^ { 2 } - 4 } { a ^ { 3 } + 4 a ^ { 2 } + 4 a } - \frac { 1 } { a } - \frac { 4 a - 4 } { a ^ { 2 } - 2 a - 8 }$$

Evaluate

$$-\frac{3}{a+2}$$

Short Solution Steps

Factor a^{3}-12a-16. Factor a^{3}+4a^{2}+4a.

$$\frac{4a^{2}+2a}{\left(a-4\right)\left(a+2\right)^{2}}-\frac{2a^{2}-4}{a\left(a+2\right)^{2}}-\frac{1}{a}-\frac{4a-4}{a^{2}-2a-8}$$

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

$$\frac{\left(4a^{2}+2a\right)a}{a\left(a-4\right)\left(a+2\right)^{2}}-\frac{\left(2a^{2}-4\right)\left(a-4\right)}{a\left(a-4\right)\left(a+2\right)^{2}}-\frac{1}{a}-\frac{4a-4}{a^{2}-2a-8}$$

Since \frac{\left(4a^{2}+2a\right)a}{a\left(a-4\right)\left(a+2\right)^{2}} and \frac{\left(2a^{2}-4\right)\left(a-4\right)}{a\left(a-4\right)\left(a+2\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(4a^{2}+2a\right)a-\left(2a^{2}-4\right)\left(a-4\right)}{a\left(a-4\right)\left(a+2\right)^{2}}-\frac{1}{a}-\frac{4a-4}{a^{2}-2a-8}$$

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

$$\frac{4a^{3}+2a^{2}-2a^{3}+8a^{2}+4a-16}{a\left(a-4\right)\left(a+2\right)^{2}}-\frac{1}{a}-\frac{4a-4}{a^{2}-2a-8}$$

Combine like terms in 4a^{3}+2a^{2}-2a^{3}+8a^{2}+4a-16.

$$\frac{2a^{3}+10a^{2}+4a-16}{a\left(a-4\right)\left(a+2\right)^{2}}-\frac{1}{a}-\frac{4a-4}{a^{2}-2a-8}$$

Factor the expressions that are not already factored in \frac{2a^{3}+10a^{2}+4a-16}{a\left(a-4\right)\left(a+2\right)^{2}}.

$$\frac{2\left(a-1\right)\left(a+2\right)\left(a+4\right)}{a\left(a-4\right)\left(a+2\right)^{2}}-\frac{1}{a}-\frac{4a-4}{a^{2}-2a-8}$$

Cancel out a+2 in both numerator and denominator.

$$\frac{2\left(a-1\right)\left(a+4\right)}{a\left(a-4\right)\left(a+2\right)}-\frac{1}{a}-\frac{4a-4}{a^{2}-2a-8}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of a\left(a-4\right)\left(a+2\right) and a is a\left(a-4\right)\left(a+2\right). Multiply \frac{1}{a} times \frac{\left(a-4\right)\left(a+2\right)}{\left(a-4\right)\left(a+2\right)}.

$$\frac{2\left(a-1\right)\left(a+4\right)}{a\left(a-4\right)\left(a+2\right)}-\frac{\left(a-4\right)\left(a+2\right)}{a\left(a-4\right)\left(a+2\right)}-\frac{4a-4}{a^{2}-2a-8}$$

Since \frac{2\left(a-1\right)\left(a+4\right)}{a\left(a-4\right)\left(a+2\right)} and \frac{\left(a-4\right)\left(a+2\right)}{a\left(a-4\right)\left(a+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2\left(a-1\right)\left(a+4\right)-\left(a-4\right)\left(a+2\right)}{a\left(a-4\right)\left(a+2\right)}-\frac{4a-4}{a^{2}-2a-8}$$

Do the multiplications in 2\left(a-1\right)\left(a+4\right)-\left(a-4\right)\left(a+2\right).

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

Combine like terms in 2a^{2}+8a-2a-8-a^{2}-2a+4a+8.

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

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

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

Cancel out a in both numerator and denominator.

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

Factor a^{2}-2a-8.

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

Since \frac{a+8}{\left(a-4\right)\left(a+2\right)} and \frac{4a-4}{\left(a-4\right)\left(a+2\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in a+8-\left(4a-4\right).

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

Combine like terms in a+8-4a+4.

$$\frac{-3a+12}{\left(a-4\right)\left(a+2\right)}$$

Factor the expressions that are not already factored in \frac{-3a+12}{\left(a-4\right)\left(a+2\right)}.

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

Extract the negative sign in 4-a.

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

Cancel out a-4 in both numerator and denominator.

$$\frac{-3}{a+2}$$

Expand

$$-\frac{3}{a+2}$$

Short Solution Steps

Factor a^{3}-12a-16. Factor a^{3}+4a^{2}+4a.

$$\frac{4a^{2}+2a}{\left(a-4\right)\left(a+2\right)^{2}}-\frac{2a^{2}-4}{a\left(a+2\right)^{2}}-\frac{1}{a}-\frac{4a-4}{a^{2}-2a-8}$$

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

$$\frac{\left(4a^{2}+2a\right)a}{a\left(a-4\right)\left(a+2\right)^{2}}-\frac{\left(2a^{2}-4\right)\left(a-4\right)}{a\left(a-4\right)\left(a+2\right)^{2}}-\frac{1}{a}-\frac{4a-4}{a^{2}-2a-8}$$

Since \frac{\left(4a^{2}+2a\right)a}{a\left(a-4\right)\left(a+2\right)^{2}} and \frac{\left(2a^{2}-4\right)\left(a-4\right)}{a\left(a-4\right)\left(a+2\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(4a^{2}+2a\right)a-\left(2a^{2}-4\right)\left(a-4\right)}{a\left(a-4\right)\left(a+2\right)^{2}}-\frac{1}{a}-\frac{4a-4}{a^{2}-2a-8}$$

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

$$\frac{4a^{3}+2a^{2}-2a^{3}+8a^{2}+4a-16}{a\left(a-4\right)\left(a+2\right)^{2}}-\frac{1}{a}-\frac{4a-4}{a^{2}-2a-8}$$

Combine like terms in 4a^{3}+2a^{2}-2a^{3}+8a^{2}+4a-16.

$$\frac{2a^{3}+10a^{2}+4a-16}{a\left(a-4\right)\left(a+2\right)^{2}}-\frac{1}{a}-\frac{4a-4}{a^{2}-2a-8}$$

Factor the expressions that are not already factored in \frac{2a^{3}+10a^{2}+4a-16}{a\left(a-4\right)\left(a+2\right)^{2}}.

$$\frac{2\left(a-1\right)\left(a+2\right)\left(a+4\right)}{a\left(a-4\right)\left(a+2\right)^{2}}-\frac{1}{a}-\frac{4a-4}{a^{2}-2a-8}$$

Cancel out a+2 in both numerator and denominator.

$$\frac{2\left(a-1\right)\left(a+4\right)}{a\left(a-4\right)\left(a+2\right)}-\frac{1}{a}-\frac{4a-4}{a^{2}-2a-8}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of a\left(a-4\right)\left(a+2\right) and a is a\left(a-4\right)\left(a+2\right). Multiply \frac{1}{a} times \frac{\left(a-4\right)\left(a+2\right)}{\left(a-4\right)\left(a+2\right)}.

$$\frac{2\left(a-1\right)\left(a+4\right)}{a\left(a-4\right)\left(a+2\right)}-\frac{\left(a-4\right)\left(a+2\right)}{a\left(a-4\right)\left(a+2\right)}-\frac{4a-4}{a^{2}-2a-8}$$

Since \frac{2\left(a-1\right)\left(a+4\right)}{a\left(a-4\right)\left(a+2\right)} and \frac{\left(a-4\right)\left(a+2\right)}{a\left(a-4\right)\left(a+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2\left(a-1\right)\left(a+4\right)-\left(a-4\right)\left(a+2\right)}{a\left(a-4\right)\left(a+2\right)}-\frac{4a-4}{a^{2}-2a-8}$$

Do the multiplications in 2\left(a-1\right)\left(a+4\right)-\left(a-4\right)\left(a+2\right).

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

Combine like terms in 2a^{2}+8a-2a-8-a^{2}-2a+4a+8.

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

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

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

Cancel out a in both numerator and denominator.

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

Factor a^{2}-2a-8.

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

Since \frac{a+8}{\left(a-4\right)\left(a+2\right)} and \frac{4a-4}{\left(a-4\right)\left(a+2\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in a+8-\left(4a-4\right).

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

Combine like terms in a+8-4a+4.

$$\frac{-3a+12}{\left(a-4\right)\left(a+2\right)}$$

Factor the expressions that are not already factored in \frac{-3a+12}{\left(a-4\right)\left(a+2\right)}.

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

Extract the negative sign in 4-a.

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

Cancel out a-4 in both numerator and denominator.

$$\frac{-3}{a+2}$$