Solve 2b/b^2+10b+16+7/b^2-64

$$\frac{ 2b }{ { b }^{ 2 } +10b+16 } + \frac{ 7 }{ { b }^{ 2 } -64 }$$

Evaluate

$$\frac{2b^{2}-9b+14}{\left(b+2\right)\left(b^{2}-64\right)}$$

Short Solution Steps

Factor b^{2}+10b+16. Factor b^{2}-64.

$$\frac{2b}{\left(b+2\right)\left(b+8\right)}+\frac{7}{\left(b-8\right)\left(b+8\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(b+2\right)\left(b+8\right) and \left(b-8\right)\left(b+8\right) is \left(b-8\right)\left(b+2\right)\left(b+8\right). Multiply \frac{2b}{\left(b+2\right)\left(b+8\right)} times \frac{b-8}{b-8}. Multiply \frac{7}{\left(b-8\right)\left(b+8\right)} times \frac{b+2}{b+2}.

$$\frac{2b\left(b-8\right)}{\left(b-8\right)\left(b+2\right)\left(b+8\right)}+\frac{7\left(b+2\right)}{\left(b-8\right)\left(b+2\right)\left(b+8\right)}$$

Since \frac{2b\left(b-8\right)}{\left(b-8\right)\left(b+2\right)\left(b+8\right)} and \frac{7\left(b+2\right)}{\left(b-8\right)\left(b+2\right)\left(b+8\right)} have the same denominator, add them by adding their numerators.

$$\frac{2b\left(b-8\right)+7\left(b+2\right)}{\left(b-8\right)\left(b+2\right)\left(b+8\right)}$$

Do the multiplications in 2b\left(b-8\right)+7\left(b+2\right).

$$\frac{2b^{2}-16b+7b+14}{\left(b-8\right)\left(b+2\right)\left(b+8\right)}$$

Combine like terms in 2b^{2}-16b+7b+14.

$$\frac{2b^{2}-9b+14}{\left(b-8\right)\left(b+2\right)\left(b+8\right)}$$

Expand \left(b-8\right)\left(b+2\right)\left(b+8\right).

$$\frac{2b^{2}-9b+14}{b^{3}+2b^{2}-64b-128}$$

Differentiate w.r.t. b

$$\frac{2\left(1024-284b-76b^{2}+9b^{3}-b^{4}\right)}{\left(\left(b+2\right)\left(b^{2}-64\right)\right)^{2}}$$