Question

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

Answer 1

15

Explanation:

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

`\frac{a}{b} xx \frac{c}{d} = \frac{ac}{bd}`

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

`\frac{3x-15}{4x^2-2x} xx \frac{10x-20x^2}{5-x} = \frac{(3x-15)(10x-20x^2)}{(4x^2-2x)(5-x)}`

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

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

`\frac{(-3)(5-x)(10x-20x^2)}{(4x^2-2x)(5-x)} = \frac{(-3)(10x-20x^2)}{(4x^2-2x)}`

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

`\frac{(-3)(-5)(4x^2-2x)}{(4x^2-2x)} = \frac{(-3)(-5)}{1} = -3xx-5 = 15`