How do you simplify #(8mn^2)/(4m)#?

1 Answer
May 14, 2015

You can see clearly that you have #m# both in the numerator and denominator of your fucntion. You can simply divide the upper one by the lower one, which will result in #1#, cancelling both of them on your function. You'll get, then, the following:

#(8*cancel(m)*n^2)/(4cancel(m)) = (8n^2)/(4)#

Now, you can divide #8/4 = 2#.

The final answer, then, will be #2n^2#.

Another possible way to adress this problem is by factoring the elements that appear up and down your function, as shown:

#(2*4*m*n^2)/(4*m)#

And now cancel the ones that appear exactly the same up and down because they do not change the value of your function, as they both multiply and divide by the same number:

#(2*cancel(4*m)*n^2)/cancel(4*m)#