How do you simplify #(12n^2+15n)/(3n)#?

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Answer 1 · 2018-03-02T04:03:19

#4n+5#

We have the function #(12n^2+15n)/(3n)#

Taking a good look at the numerator, we realize that we can factor #3n# out:

#(3n(4n+5))/(3n)#

We have a #3n# on top and on the bottom, so this becomes:

#4n+5#, the answer!

Answer 2 · 2018-03-02T04:06:09

You have to find the factors of all of the terms, then simplify as needed.

You start off with this:

#(12n^2 + 15n)/(3n)#

Then, you take out the common factors:

#(3n(4n+5))/(3n)#

Finally, you simplify by cancelling the #3n# from the numerator and the denominator:

#4n+5#

And you're done!