How do you divide {(n^2-n-12)/(2n^2-15n+18)}/{(3n^2-12n)/(2n^3-9n^2)}?

Apr 3, 2016

$\frac{n \left(n + 3\right) \left(2 n - 9\right)}{3 \left(2 n - 3\right) \left(n - 6\right)}$

Explanation:

$\frac{\frac{{n}^{2} - n - 12}{2 {n}^{2} - 15 n + 18}}{\frac{3 {n}^{2} - 12 n}{2 {n}^{3} - 9 {n}^{2}}}$ is equivalent to $\frac{{n}^{2} - n - 12}{2 {n}^{2} - 15 n + 18} \times \frac{2 {n}^{3} - 9 {n}^{2}}{3 {n}^{2} - 12 n}$, as a division by a fraction is equivalent to multiplication by its multiplicative inverse.

For solving the above we need to factorize each polynomial.

n^2-n-12=(n^2-4n+3n-12=n(n-4)+3(n-4)=(n+3)(n-4)

$2 {n}^{2} - 15 n + 18 = 2 {n}^{2} - 12 n - 3 n + 18 = 2 n \left(n - 6\right) - 3 \left(n - 6\right) = \left(2 n - 3\right) \left(n - 6\right)$

$3 {n}^{2} - 12 n = 3 n \left(n - 4\right)$ and $2 {n}^{3} - 9 {n}^{2} = {n}^{2} \left(2 n - 9\right)$

Hence, $\frac{{n}^{2} - n - 12}{2 {n}^{2} - 15 n + 18} \times \frac{2 {n}^{3} - 9 {n}^{2}}{3 {n}^{2} - 12 n}$ is equivalent to

$\frac{\left(n + 3\right) \cancel{\left(n - 4\right)}}{\left(2 n - 3\right) \left(n - 6\right)} \times \frac{n \left(\cancel{{n}^{2}}\right) \left(2 n - 9\right)}{3 \cancel{n} \cancel{\left(n - 4\right)}}$

= $\frac{n \left(n + 3\right) \left(2 n - 9\right)}{3 \left(2 n - 3\right) \left(n - 6\right)}$