How do you divide #(3a^3+17a^2+12a-5)/(a+5)#?

1 Answer
John D.
Jul 14, 2017

#3a^2+2a+2 - 15/(a+5)#

Explanation:

First, let's split the top polynomial up into "multiples" of the bottom polynomial. To see what I mean, let's first try to take care of the #3a^3# term.

We can see that #3a^2(a+5) = 3a^3+15a^2#. So, let's separate this from our polynomial:

#3a^3 + 17a^2 + 12a - 5#

#(3a^3+15a^2) + 2a^2+12a-5#

#3a^2(a+5) + 2a^2+12a-5#

See how this "gets rid of" the #3a^3# term? Let's do the same for the #2a^2# term.

We can see that #2a(a+5) = 2a^2 + 10a#.

#3a^2(a+5) + (2a^2+10a) + 2a-5#

#3a^2(a+5) + 2a(a+5)+ 2a-5#

The next highest term to deal with is the #2a# term.

We can see that #2(a+5) = 2a+10#.

#3a^2(a+5) + 2a(a+5)+ (2a+10) -15#

#3a^2(a+5) + 2a(a+5)+ 2(a+5) - 15#

We can't do anything about the #15#, since its degree is smaller than the degree of #a+5#.

Finally, let's divide everything by #(a+5)#. This is why we wrote everything in terms of #a+5#, so we can just cancel out the top and the bottom of the fraction at this step.

#(3a^3+17a^2+12a-5)/(a+5) #

#= (3a^2(a+5) + 2a(a+5)+ 2(a+5) - 15)/(a+5)#

#= 3a^2+2a+2 - 15/(a+5)#

Final Answer

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