Solve 4a^2+4a+1/(a+6)^2div6a^2-5a-4/3a^2+14a-24

$$\frac { 4 a ^ { 2 } + 4 a + 1 } { ( a + 6 ) ^ { 2 } } \div \frac { 6 a ^ { 2 } - 5 a - 4 } { 3 a ^ { 2 } + 14 a - 24 }$$

Evaluate

$$\frac{2a+1}{a+6}$$

Short Solution Steps

Divide \frac{4a^{2}+4a+1}{\left(a+6\right)^{2}} by \frac{6a^{2}-5a-4}{3a^{2}+14a-24} by multiplying \frac{4a^{2}+4a+1}{\left(a+6\right)^{2}} by the reciprocal of \frac{6a^{2}-5a-4}{3a^{2}+14a-24}.

$$\frac{\left(4a^{2}+4a+1\right)\left(3a^{2}+14a-24\right)}{\left(a+6\right)^{2}\left(6a^{2}-5a-4\right)}$$

Factor the expressions that are not already factored.

$$\frac{\left(3a-4\right)\left(a+6\right)\left(2a+1\right)^{2}}{\left(3a-4\right)\left(2a+1\right)\left(a+6\right)^{2}}$$

Cancel out \left(3a-4\right)\left(a+6\right)\left(2a+1\right) in both numerator and denominator.

$$\frac{2a+1}{a+6}$$

Expand

$$\frac{2a+1}{a+6}$$

Short Solution Steps

Divide \frac{4a^{2}+4a+1}{\left(a+6\right)^{2}} by \frac{6a^{2}-5a-4}{3a^{2}+14a-24} by multiplying \frac{4a^{2}+4a+1}{\left(a+6\right)^{2}} by the reciprocal of \frac{6a^{2}-5a-4}{3a^{2}+14a-24}.

$$\frac{\left(4a^{2}+4a+1\right)\left(3a^{2}+14a-24\right)}{\left(a+6\right)^{2}\left(6a^{2}-5a-4\right)}$$

Factor the expressions that are not already factored.

$$\frac{\left(3a-4\right)\left(a+6\right)\left(2a+1\right)^{2}}{\left(3a-4\right)\left(2a+1\right)\left(a+6\right)^{2}}$$

Cancel out \left(3a-4\right)\left(a+6\right)\left(2a+1\right) in both numerator and denominator.

$$\frac{2a+1}{a+6}$$