Expert: Scott A Wilson - 1/13/2010

Question
How do I divide (5a^3+2)/(3a+6) and (4a^2-a)/(3a+6).  Is there a way to divide them.  I am trying to help figure out some function problems (f(a)/g(a)).  I hope I have the problems right.

Answer
It is similar to long division by hand.
Instead of 1's, 10's, and 100's, we have 1's, a's, and a²'s.
The first one involes dividing 3 6 into 5 0 2.
The 0 was put in since there are no a's.

For the first one, we have (3 6) divided into (5 0 2).
Looking at the start of each set, the number to multiply by is 5/3.
We put a 5/3 on the top and subtract off (5/3)(3 6).
Now (5/3)(3 6) is (5 10).  Subtracting that off of the first two terms of (5 0 2) gives (0 -10 2)

(3 6) divided into (-10 2) is -10/3.  Putting the -10/3 at the top, we then subtract off
-10/3 times (3 6).  Subtracting a negative is the same as addding.
That is (-10,2) + (10/3)(3 6) = (-10,2) + (10 20) = (0 22).  This is the remainder, 22.

So we have (3a+6)(5a/3 - 10/3) + 22 = 5a² (a's cancel) -20 + 22 = 5a² + 2.


In a similar fashion, we can do (3 6) into (4 -1 0).
Multiply by 4/3 and subtract.
(4/3)(3 6) = 4 8.
(4 -1 0) gives us a (4 -1) in fron, and (4 -1) - (4 8) gives us (0 7)
Bringing down the 0 at the end and dropping the 0 at the start to (0 7) gives us (7 0).

Here, we need to multiply by 7/3.
That gives us (7/3)(3 6) = (7 14).
Now (7 0) - (7 14) gives (0 -14), so - 14 is the remainder.
So the answer would be (4a²-a)/(3a+6) = (4a/3 + 7/3)(3a+6) - 14.
That works out to 4a² - a + 14 - 14 = 4a² - a.