Partial FractionsDate: 11/23/2003 at 16:09:41 From: jamie Subject: Partial Fractions I need to put this fraction into power series formation: 3x^2 - x ----------------- x^3 - x^2 - x + 1 I'm not sure how to set up partial fractions because it doesn't factor. The farthest I get on factoring is this: 3x^2 - x ------------------- x(x^2 - x - 1) + 1 Date: 11/23/2003 at 22:33:00 From: Doctor Jordan Subject: Re: Partial Fractions Dear Jamie, Your thinking about wanting to factor is correct, and you are getting stuck because you have not been able to factor the denominator, which can actually be factored. Let's look at how. By observation we see that x^3 - x^2 - x + 1 has a value of zero for x = 1. So we know that (x - 1) is a factor of it. Using long or synthetic division with (x - 1), we find that it factors into: (x - 1)(x^2 - 1) then noting that the second piece is a difference of squares we get (x - 1)(x + 1)(x - 1) which can be written (x - 1)^2(x + 1) So we set up our partial fractions like this: 3x^2-x A B C ----------- = ------ + ----- + ----- (x-1)^2(x+1) (x-1)^2 (x-1) (x+1) We see with the A and B fractions that if we have a factor that is squared, we must have a variable over each of its decreasing powers, from its original power through to 1 (that is, we have A over the factor to the power of 2 [its original power], and B over the factor to the power of 1). We can now solve this problem; we multiply A by (x+1), B by (x-1) (x+1), and C by (x-1)^2. Doing this achieves a common denominator on the right hand side of (x-1)^2(x+1), and since we have the same denominator on the left and right sides, we can multiply both sides by that denominator to eliminate it. Then we proceed to substitute in x values, solving this equation the standard way for partial fractions. If you have any other questions, or didn't quite like this explanation, please write me back. Good luck! - Doctor Jordan, The Math Forum http://mathforum.org/dr.math/ |
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