Solve a^2-a-12/2a^2+adiv16-a^2/2a^2+9a+4

$$\frac { a ^ { 2 } - a - 12 } { 2 a ^ { 2 } + a } \div \frac { 16 - a ^ { 2 } } { 2 a ^ { 2 } + 9 a + 4 }$$

Evaluate

$$-1-\frac{3}{a}$$

Short Solution Steps

Divide \frac{a^{2}-a-12}{2a^{2}+a} by \frac{16-a^{2}}{2a^{2}+9a+4} by multiplying \frac{a^{2}-a-12}{2a^{2}+a} by the reciprocal of \frac{16-a^{2}}{2a^{2}+9a+4}.

$$\frac{\left(a^{2}-a-12\right)\left(2a^{2}+9a+4\right)}{\left(2a^{2}+a\right)\left(16-a^{2}\right)}$$

Factor the expressions that are not already factored.

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

Extract the negative sign in 4+a.

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

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

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

Expand the expression.

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

Expand

$$-1-\frac{3}{a}$$

Short Solution Steps

Divide \frac{a^{2}-a-12}{2a^{2}+a} by \frac{16-a^{2}}{2a^{2}+9a+4} by multiplying \frac{a^{2}-a-12}{2a^{2}+a} by the reciprocal of \frac{16-a^{2}}{2a^{2}+9a+4}.

$$\frac{\left(a^{2}-a-12\right)\left(2a^{2}+9a+4\right)}{\left(2a^{2}+a\right)\left(16-a^{2}\right)}$$

Factor the expressions that are not already factored.

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

Extract the negative sign in 4+a.

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

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

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

Expand the expression.

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