more partial fractions?

Question

a) (x^2 - x - 1) / [ x(x+1)(x-1)


b) 3 / (x+1)(x+2)(x+3)



c) (x^2 + 10x - 6) / [x^2 (x-1) ]



d) (x^2 - 3x + 1) / (x-1)^3





Thanks sooo much!!
♥

sorry, but i was looking for some working, at least for one, i'm not just copying these for homework. I need to figure out how to do them right!!
eeeek!

Answer 1

a. 1/x + 0.5/(x+1) - 0.5/(x-1)


b. 3/(2*(1 + x)) - 3/(2 + x) + 3/(2*(3 + x))

§
c. 5/(x - 1) + 6/x^2 - 4/x


d. -(x - 1)^(-3) - (x - 1)^(-2) + (x - 1)^(-1)

Answer 2

For the examples where there are no multiple roots (which is to say, none of the factors in the denominator is raised to a power greater than 1), it is pretty easy.

a) (x^2 - x - 1) / [x(x + 1)(x - 1)]

The "poles" (values for which the denominator is equal to zero) are 0, -1, and 1. We call them p1 = 0, p2 = -1, and p3 = 1. It doesn't matter which order you assign.

We write (x^2 - x - 1) / [x(x + 1)(x - 1)] = a / x + b / (x + 1) + c / (x -1). And now, we use a simple process. To find the numerator that corresponds to a denominator of (x - pn), where pn is any of the poles, you simply evaluate the original fraction multiplied by (x - pn) for x = pn.

For example, a corresponds to the denominator x, for p1 = 0. So a = (x)(x^2 - x - 1) / [x(x + 1)(x - 1)] = (x^2 - x - 1) / [(x + 1)(x - 1)] evaluated at x = 0, which is (-1) / [(1)(-1)] = -1 / -1 = 1. Continuing, b = (x^2 - x - 1) / [x(x - 1)] evaluated at x = -1, which is (1 + 1 - 1) / [(-1)(-2)] = 1/2, and c = (x^2 - x - 1) / [x(x + 1)] evaluated at x = 1, which is (1 - 1 - 1) / [(1)(2)] = -1/2.

So (x^2 - x - 1) / [x(x + 1)(x - 1)] = 1 / x + (1/2) / (x + 1) - (1/2) / (x - 1).

To understand why this works, look at the original partial fraction expansion, before we knew what a, b, and c were. If x = pn, then x - pn = 0. So multiplying the expansion through by (x - pn) when x = pn causes the fractions without (x - pn) in their denominators to go to zero, while the term that does have (x - pn) in the denominator becomes equal to its numerator (because the zero values get cancelled out). And since the expansion is equal to the original fraction, you can use the fraction to find the value.

It is more difficult in a case like d), where there is a multiple root.

d) (x^2 - 3x + 1) / (x - 1)^3

You have to deal with each power of the denominator, so your expansion is a / (x - 1)^3 + b / (x - 1)^2 + c / (x - 1). You can find a the same way as above, by taking the value of (x - 1)^3 times the original fraction, which would be x^2 - 3x + 1, evaluated at x = 1. I get a = -1. You can find b by using differentiation. After mulitplying through by (x - 1)^3, you have x^2 - 3x + 1 = a + b(x - 1) + c(x - 1)^2. If you differentiate both sides, you get 2x - 3 = b + 2c(x - 1). Evaluating at x = 1 gives you -1 = b. Differentiating again gives you 2 = 2c ==> c = 1. (Trivially, this is also evaluated at x = 1, but it doesn't matter because x did not appear in the expression.)

So (x^2 - 3x + 1) / (x - 1)^3 = -1 / (x - 1)^3 - 1 / (x - 1)^2 + 1 / (x - 1).

In the case of (c), which contains both types of poles, break it down into a / (x - 1) + B / x^2, where B is not a single number but rather an unknown expression. Start by using the first method to find the numerator that goes with (x - 1). Then, once you know the value of a / (x - 1), use algebra to find the value of B, which will probably be of the form cx + d. Then use the second method to break down (cx + d) / x^2 as if it were a new partial fraction problem.