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

1 Answer
Jacob
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:

#\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#

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