long division of rational functions?

Question

i'm in AP calculus and have now come to the realization that i can't use long division to solve rational functions......i can use synthetic division....but that won't work for all rational functions.....

f(x)=(x^3)/(4-x^2)


here's my work so far.....
-x^2+4 / x^3 + 0x^2 + 0x

-1x
--------------------
-x^2+4 / x^3 + 0x^2 + 0x
x^3 -4x
---------------------
x^2+4x

ok, the math up there is hard to read so let me clarify what i did....first i took -1x and multiplied that by -x^2-4....which left me with 0x^2 + 4x
...but i don't see how i can continue after that....any insight would be appreciated

Answer 1

Why are you doing the division anyway? What are you trying to do with the function?

If you want to graph it, you should recognize by inspection that the fuction has vertical asymptotes at 2 and -2 (the zeroes of the denominator). It has a zero when x = zero, which is the only place it crosses the x axis. It changes signs at that point. It also changes signs when you cross the asymptotes, becoming negative when x>2 and becoming positive when x<-2. Looking at the dominant terms of the numerator and denominator, for large positive values of x, this function is approximately x^3/(-x^2) which reduces to -x, which means as x=>∞, f(x)=>-∞ and as x=>-∞, f(x)=>∞.

If you want to find the derivative of this function, you use the quotient rule, so

f'(x) = [(4-x^2)3x^2-x^3(-2x)]/[(4-x^2)^2] which can be simplified.

There are also techniques for integrating this kind of function, which I don't want to get into here, but since it's early in the year, I'm guessing you're not doing those yet.

So I'm not getting why you want to use division. There is nothing to "solve" here, only a function which can be understood and whose behavior can be described.

Answer 2

You have a couple of options. One is to just interpret that 4x as the remainder, like others have said.

Another possibility is to develop and infinite expansion by dividing 4x by 4-x^2 and getting (-4/x). When you multiply out and find the new remainder (which is -16/x), you divide that by 4-x^2 to get (-16/x^3) with a remainder of 64/x^3. Keep going. This is very similar to dividing 10 by 3. You will get either 3 with a remainder of 1 OR 3.3333.....

Answer 3

This is the p[oints of stop. You have done nothing wrong. Only one thing thay you should realise that there can be a remainder just like in arithemtic(h

x^3/(4-x^2) = -(x) + 4x/(4-x^2)

we can cross check by multiplying
-(4-x^2)(x)
= (x^2-4)(x) = x^3 - 4x

add 4x and we get the result
The quoteint = -x
ramainder = 4x

Answer 4

..........-x
-x²+4 | x³
.......... x³ ......-4x
..........--------------
.....................4x

Thus your result is -x + 4x/(-x²+4). Part of the reason you're confused is because in this problem, you can only perform one step before ending up with a remainder. As such, it might seem like you're doing something wrong. You didn't - once you get to the point where the highest remaining term, all that's left to do is write the result. Depending on the problem you're trying to solve, you may want to start applying partial fractions to the remainder at this point.

Answer 5

You can't continue if the degree of the remainder is less than the degree of the divisor. The final answer is -x + 4x/(4-x^2)

Answer 6

-x^2+0x+4)x^3 +0x^2+0x+0(-x
...............-(x^3 +0x^2-4x)
.......................................
............................4x (remainder)
Therefore

f(x)=(x^3)/(4-x^2)= - x +4x/(4-x^2)

Answer 7

What are you trying to do with the function?