# How do you multiply (3m^2-12m)div(m^2-4m)/(m^2-6m+8)?

Jul 2, 2017

$3 {m}^{2} - 18 m + 24$

#### Explanation:

Let's factorise everything we've got:

$\frac{3 {m}^{2} - 12 m}{1} \div \frac{{m}^{2} - 4 m}{{m}^{2} - 6 m + 8}$

When we are dividing fractions, we can multiply by the reciprocal of the second fraction, so our expression really is

$\frac{3 {m}^{2} - 12 m}{1} \times \frac{{m}^{2} - 6 m + 8}{{m}^{2} - 4 m}$

Let's factorise everything we can:

$\frac{3 m \left(m - 4\right)}{1} \times \frac{\left(m - 2\right) \left(m - 4\right)}{m \left(m - 4\right)}$

Now, let's see what cancels out, meaning what has the same factor in the numerator as the denominator

$\frac{3 \cancel{m} \cancel{\left(m - 4\right)}}{1} \times \frac{\left(m - 2\right) \left(m - 4\right)}{\cancel{m} \cancel{\left(m - 4\right)}}$

Now let's multiply the fractions. Remember, it's straight across!

$\frac{3 \times \left(m - 2\right) \left(m - 4\right)}{1}$

$3 \times \left({m}^{2} - 6 m + 8\right)$

$3 {m}^{2} - 18 m + 24$