Calculus (Partial Fractions)
posted by Sean on .
My Calculus class just started Partial Fractions, and I understand MOST of it.
I'm having a slight problem though, when either C, or C and D come in.
A/(something) + B/(something)
But I become confused when it's either
A/(something) + B/(something) + C/(something)
A/(something) + B/(something) + (Cx + D)/(something)
Why is it that you get ONLY C, or Cx + D ?
I always get confused when it comes to that, so help would be greatly appreciated.
When you have a quadratic that does not have real roots in the denominator (complex roots only), you put Cx+D in the numerator.
like if your denominator were (x-1)(x+1)(x-2)
you might use
A/(x-1) + B/(x+1) +C/(x-2)
But if your denominator were:
You could deal with the (x-1) part just fine.
but what to do with the other part?
You can not write it as the sum of two fractions with numerators B and C because you can nor write
x^2-2x+2 as (x+p)(x+q)
so you have to resort to
(C + D x)/(x^2-2x+2)
This method has practical applications only if the denominator factors.
Since you don't give an example I will supply one
(I started with known fractions and simplified, so that I would have a question that worked out)
The bottom factors to x(x+2)(x-1)
(5x^2+3x+4)/(x^3+x^2-2x) = A/(x+2) + B/(x-1) + c/x
= [Ax(x-1) + Bx(x+2) + C(x+2)(x-1)]/x(x+2)(x-1)
clearing the denominator we get:
5x^2+3x+4 = Ax(x-1) + Bx(x+2) + C(x+2)(x-1)
now let x=0, then -2C = 4, and C = -2
let x=1, then 3B=12, and B = 4
let x=-2, then 6A = 18 and A = 3
so my original fraction (5x^2+3x+4)/(x^3+x^2-2x) can
be split into
3/(x+2) + 4/(x-1) - 2/x
Reiny. You misunderstood. And I didn't word it right.
I meant where does (Cx + D)/Something
If the Denominator is
x^4 - 2x^2 - 8 > (x-2) (x+2) (x^2+2)
Would I do A + B + C, or A + B + (Cx + D)?
Wow. I'm making this topic last a while.
I think- and this is just out loud- that you would do (Cx + D) Over the X^2 + 2?
Am I getting that right?
You have Cx + D over something if the SOMETHING has a term of X to a power greater than one?