Hey guys i'm a bit confused and clueless when it comes to writing out the form of the partial fraction decomposition of the function.
Maybe you guys can help me understand through examples.
2 / (x^2 + 3x - 4)
now I know I can factor the Q(x) to (x + 4)(x - 1)
So the partial fractioning for this will be (A / (x + 4)) + (B / (x - 1)) right? I have no problems doing simple ones such as this.
Now here is an example of one of which i'm totally clueless as where the extra factors are coming from.
1 / (x^6 - x^3)
now I can factor the Q(x) to (x^3)(x - 1)(x^2 +x + 1). Now how would I partial fraction this and the most important WHY? The "WHY" is what I really need most. Thanks
When dealing with powers, for example [P(x)]^n, with P(x) a polynomial of degree 1 (that is to say x-4, 2x+3, etc...), we write this :
When you have an irreducible polynomial, such as , then :
Edit : niaaah >.< too slcow !
For repeated factors, I must include a term and it goes the same for quadratics. Let me try one on here and tell me if it's correct.
(2x + 1) / ((x + 1)^3)((x^2 + 4)^2)
(A / (x + 1)) + B / ((x + 1)^2) + C / ((x + 1)^3) + Dx +E / (x^2 + 4) + Fx + G / ((x^2 +4)^2)
what do you think?
I really appreciate the help guys! I have one more question
How am I supposed to know when to Equate for quadratics, linear, and constants?
Example. ( This is kinda ugly )
Example : (-x^3 + 2x - x + 1) = A(x^4 + 2x^2 + 1) + B(x^4 + x^2) + C(x^3 + x) + Dx^2 + Ex.
Thanks in advance
First we multiply out the right hand side to get
Now we group all terms of the same degree (same exponent) to get
Now we can factor the right hand side to get
Now we can equate the coeffients from both sides of the equation. For exapmle since there is no term on the left hand side its coeffeint must be zero so... now for the cubic term and so on and finally
Now we can solve this system to get A=1, B=-1,C=-1,D=1,E=0