# How do you divide (4k^2-29k-24)/(k^2-13k+40) by (4k^2+15k+9)/(5k^3-25k^2-3k+15)?

Dec 4, 2016

$\frac{\left(5 {k}^{2} - 3\right)}{\left(k + 3\right)}$

#### Explanation:

With algebraic fractions, the first approach is to factorise where possible:

$\frac{4 {k}^{2} - 29 k - 24}{{k}^{2} - 13 k + 40} \div \frac{4 {k}^{2} + 15 k + 9}{5 {k}^{3} - 25 {k}^{2} - 3 k + 15}$

$= \frac{\left(4 k + 3\right) \left(k - 8\right)}{\left(k - 8\right) \left(k - 5\right)} \div \frac{\left(4 k + 3\right) \left(k + 3\right)}{5 {k}^{2} \left(k - 5\right) - 3 \left(k - 5\right)}$

To divide by a fraction, multiply by the reciprocal.

$= \frac{\left(4 k + 3\right) \left(k - 8\right)}{\left(k - 8\right) \left(k - 5\right)} \times \frac{\left(k - 5\right) \left(5 {k}^{2} - 3\right)}{\left(4 k + 3\right) \left(k + 3\right)}$

Now cancel like factors

$= \frac{\cancel{\left(4 k + 3\right)} \cancel{\left(k - 8\right)}}{\cancel{\left(k - 8\right)} \cancel{\left(k - 5\right)}} \times \frac{\cancel{\left(k - 5\right)} \left(5 {k}^{2} - 3\right)}{\cancel{\left(4 k + 3\right)} \left(k + 3\right)}$

$= \frac{\left(5 {k}^{2} - 3\right)}{\left(k + 3\right)}$