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.
I understand
A/(something) + B/(something)
But I become confused when it's either
A/(something) + B/(something) + C/(something)
OR
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:
(x-1)(x^2-2x+2)
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
separate (5x^2+3x+4)/(x^3+x^2-2x)
(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)
so let
(5x^2+3x+4)/(x^3+x^2-2x) = A/(x+2) + B/(x-1) + c/x
(5x^2+3x+4)/(x^3+x^2-2x)
= [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
Pop up?
Okay. Example.
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?
OR
You have Cx + D over something if the SOMETHING has a term of X to a power greater than one?
I can understand how you might find it confusing when dealing with partial fractions involving multiple terms like A/(something) + B/(something) + C/(something) or A/(something) + B/(something) + (Cx + D)/(something). Let me explain the reason for this confusion.
In partial fractions, the general approach is to break down a complicated fraction into simpler fractions. When you have a denominator of (something) repeated multiple times in the original fraction, you need to use a combination of different terms to represent it.
Let's take the first case: A/(something) + B/(something) + C/(something). Here, (something) is repeated three times in the denominator. So, you need three different terms to represent it. Each term contributes its own constant (A, B, and C) divided by (something).
Now let's move on to the second case: A/(something) + B/(something) + (Cx + D)/(something). In this case, there is an additional term (Cx + D) in the numerator, which is a linear expression of the variable x. When you have a linear expression in the numerator, you need to break it down into separate terms using the same denominator. So, you combine the constants (C and D) and the linear expression (Cx + D) to get (Cx + D)/(something). Note that you do not separate C and D into two separate terms because they are part of the same linear expression.
To find the values of A, B, C, and D, you need to equate the numerators of both sides of the equation and compare the coefficients of the corresponding powers of x. By equating the numerators, you will end up with a system of linear equations that can be solved to find the values of A, B, C, and D.
In summary, the reason for using only C or Cx + D in partial fractions is because of the way the complicated fraction is broken down into simpler fractions, taking into account the repetitive nature of the denominator or any additional linear terms in the numerator.
I hope this explanation helps clarify your confusion. Feel free to ask any further questions you may have!