Solve b^2/b^2+2b+1quad3/b^2-8b-9

$$\frac { b ^ { 2 } } { b ^ { 2 } + 2 b + 1 } \quad \frac { 3 } { b ^ { 2 } - 8 b - 9 }$$

Evaluate

$$\frac{3b^{2}}{\left(b-9\right)\left(b+1\right)^{3}}$$

Short Solution Steps

Multiply \frac{b^{2}}{b^{2}+2b+1} times \frac{3}{b^{2}-8b-9} by multiplying numerator times numerator and denominator times denominator.

$$\frac{b^{2}\times 3}{\left(b^{2}+2b+1\right)\left(b^{2}-8b-9\right)}$$

Use the distributive property to multiply b^{2}+2b+1 by b^{2}-8b-9 and combine like terms.

$$\frac{b^{2}\times 3}{b^{4}-6b^{3}-24b^{2}-26b-9}$$

Differentiate w.r.t. b

$$\frac{6b\left(-b^{2}+5b-9\right)}{\left(b-9\right)^{2}\left(b+1\right)^{4}}$$