Multiplying Rational Expressions

I am having some serious trouble. An example question that is shown to me is as follows:

The example shows this work to solve the problem:

which is then simplified to get the answer m² / 2n³

I am COMPLETELY lost as to how they factored(?) the denominator in the second expression. I have looked at multiple examples and it's just not clicking. I understand every other part of this problem and similar problems, but that bolded part has me completely stumped.

For clarity, here is another example. I am not understanding how the example is coming up with that simplification for the second denominator.


Any advice?

Thank you! Getting every problem right so far. Your explanation was MUCH clearer than the one in the book, I'll be using that method from now on.

Just out of curiousity, why was the book explaining it the way it was written in the OP? Anyone able to explain the process in the original steps?

At the very minimum use parentheses! I think you mean (3m^4n^7 / m-5n )* (2m-10n / 12m^2n^10) or, better, \(\displaystyle \frac{3m^4n^7}{m- 5n}\frac{2m- 10n}{12m^2n^{10}}\).

According to you "they" wrote that as \(\displaystyle \frac{3m^4n^7}{m- 5n}\frac{2(m- 5n)}{2(3m^2n^7)(2n^3)}\)
In the numerator of the second fraction, they used the fact that 2m- 10n= 2m- 2(5)n= 2(m- 5n) of course. And they clearly did that because that "m- 5n" will cancel with the "m- 5n" with the numerator of the first fraction.

For the denominator, they have, first, used the fact that 12= 2*2*3 and written the "2" out in front, the "3" in the middle group, and that last "2" with the third group. They have also used the fact that \(\displaystyle n^{10}= (n^7)(n^3)\). The reason to do that is to compare that term in the middle "[itex]3m^2n^7[/itex]" with the "[itex]3m^4n^7[/itex]" in the numerator of the first fraction.

For your second example, \(\displaystyle \frac{3 m^4n^7}{4 m^2n^{10}}\frac{2m- 10n}{m- 5n}\).

You can deal with those two fractions separately: \(\displaystyle \frac{3 m^4 n^7}{12m^2n^{10}}=\)\(\displaystyle \frac{3}{12}\frac{m^4}{m^2}\frac{n^7}{n^{10}}\). 12= 3(4) so \(\displaystyle \frac{3}{12}= \frac{3}{3(4)}= \frac{1}{4}\). The two "m"s in the denominator cancel two in the numerator leaving two: \(\displaystyle m^{4- 2}= m^2\). The 7 "n"s in the numerator cancel 7 in the denominator leaving three: \(\displaystyle \frac{1}{n^{10- 7}}= \frac{1}{n^3}\). So \(\displaystyle \frac{3 m^4 n^7}{12 m^2 n^{10}}= \frac{m^2}{4 n^3}\). The second fraction is \(\displaystyle \frac{2m- 10n}{m- 5n}= \frac{2(m- 5n)}{m- 5n}= 2\) because the two "m- 5n" terms cancel.