# How do you multiply (3x-15)/(4x^2-2x)*(10x-20x^2)/(5-x)?

Jun 9, 2017

15

#### Explanation:

When multiplying fractions, whether with unknowns in algebra or not, simply multiply the numerators together, and multiply the denominators together, so:

$\setminus \frac{a}{b} \times \setminus \frac{c}{d} = \setminus \frac{a c}{b d}$

In this case, remembering to keep things in brackets, that would give us:

$\setminus \frac{3 x - 15}{4 {x}^{2} - 2 x} \times \setminus \frac{10 x - 20 {x}^{2}}{5 - x} = \setminus \frac{\left(3 x - 15\right) \left(10 x - 20 {x}^{2}\right)}{\left(4 {x}^{2} - 2 x\right) \left(5 - x\right)}$

However this can be simplified by removing common factors of the top and bottom.

We can take out a factor of -3, and write $\left(3 x - 15\right)$ as $\left(- 3\right) \left(5 - x\right)$. You'll now see that there is a common factor of $5 - x$ on the top and bottom, so:

$\setminus \frac{\left(- 3\right) \left(5 - x\right) \left(10 x - 20 {x}^{2}\right)}{\left(4 {x}^{2} - 2 x\right) \left(5 - x\right)} = \setminus \frac{\left(- 3\right) \left(10 x - 20 {x}^{2}\right)}{\left(4 {x}^{2} - 2 x\right)}$

Also we can take out a factore of -5 and write $\left(10 x - 20 {x}^{2}\right)$ as $\left(- 5\right) \left(4 {x}^{2} - 2 x\right)$, so again we can cancel out the $\left(4 {x}^{2} - 2 x\right)$ on the top and bottom so:

$\setminus \frac{\left(- 3\right) \left(- 5\right) \left(4 {x}^{2} - 2 x\right)}{\left(4 {x}^{2} - 2 x\right)} = \setminus \frac{\left(- 3\right) \left(- 5\right)}{1} = - 3 \times - 5 = 15$