Advanced Math/Partial Fractions
Hey Randy, Im struggling on a few questions and was wondering if you could help me out so that i can see how it done correctly, the question is on Partial Fractions and reads.
Resolve the following into partial fractions
a) 3(2x^2-8x-1) / (x+4)(x+1)(2x-1)
b) 3x^2-8x-63 / x^2-3x-10
Any help would be appreciated!
The key to partial fractions is to write out the form of the equation you are shooting for using "undetermined coefficients" and then do algebra to solve for these coeffs.
For your example a), we want to separate the factors in the denominator into a sum of rational functions, in this case
3(2x^2-8x-1) / (x+4)(x+1)(2x-1) = A/(x+4) + B/(x+1) + C/(2x-1)
where we want to solve for A, B and C.
This is the first step. There are fairly simple rules for how to write the right hand side (RHS), for which you should consult a textbook. One easy-to-remember rule is that the degree (highest power of x) of the numerator must be less than the denominator, otherwise you need to do a long division of the polynomials to get a remainder that you can then apply the partial fraction expansion to. Note that the expression above is x^3 in the denominator (if you multiply it all out) and only x^2 in the numerator.
At any rate, the expression above will work. The next step is to multiply both sides by the denominator of the LHS. I get
6x^2 -24x -3 = A(x+1)(2x-1) + B(x+4)((2x-1) + C(x+4)(x+1).
The 3rd step is to factor out the powers of x; for the RHS I get (check my algebra!)
(x^2)･(2A +2B +c) +x･(A +7B +5C) + x^0･(-A - 4B + 4C) where x^0 = 1.
We can now obtain 3 (independent) equations for A, B and C by matching the powers of x on both sides of the equation. Technically, this step is justified because the powers-of-x terms are orthogonal (don't project on to each other) so that the coefficients must be equal on both sides. Rearranging, this gives
x^2 → -6 = 2A + 2B + C
x^1 → -24 = A + 7B + 5C
x^0 → -3 = -A - 4B + 4C.
OK, the task now is to apply your favorite technique (Gaussian elimination?) to solve for A, B and C. I'll let you take care of it.
For your 2nd case, the denominator factors into (x-5)(x+2), which would seem to allow us to apply the technique above pretty much the same way. However, the degree of the numerator and denominator are the same (= 2), so division must be performed first. Again, I'll let you try it!
Good luck and let me know how it goes.