Partial Fractions Problem

Factual Questions
1

I need to expand the fraction (6x^2-5x)/(x+2)^3. To do this my book tells me to factor the denominator and use ABC variables for the numerator, and set the sum of it equal to the original numerator:

6x^2 - 5x = A(x+2)^2(x+2)^3 + B(x+2)(x+2)^3 + C(x+2)(x+2)^2

Then they say to pick a conveinent x to solve for one of the letters. I do this for all the letters. The problem here is that I can’t cancel out any of the letters. I can’t get any of them to equal zero. I know I could solve a system of 3 equations but that is a long process and I don’t think we learned that or need to learn that from where we are in the class right now. Is there some trick im missing or some other trick I can use to solve this?

Thanks.

2

I’m not 100% on partial fractions, but I learned it like this:

(6x^2-5x)/(x+2)[sup]3[/sup] = A/(x+2) + B/(x+2)[sup]2[/sup] + C/(x+2)[sup]3[/sup]

Then multiply through by (x+2)[sup]3[/sup]. This gives a somewhat different polynomial than the one you have.

3

I recommend Achenar’s method.

But the way I would do it is factorize the numerator a term at a time.

6x[sup]2[/sup]-5x
=(6(x+2)[sup]2[/sup]-[some crap with x and constants in])-5x
=6(x+2)[sup]2[/sup]+?x+?
=6(x+2)[sup]2[/sup]+?(x+2)-?+?
=6(x+2)[sup]2[/sup]+?(x+2)+?

And then divide each term by (x+2)[sup]2[/sup]. (The ? represent some number. It should be clear which are the same. Some +s are probably -s.)

Does that help at all?

4

Achernar’s method is correct.

Multiplying through by (x+2)^3 gets you:

(6x^2 - 5x) = A(x+2)^2 + B(x+2) + C

set x = -2, and then C = 34

(6x^2 - 5x - 34) = A(x+2)^2 + B(x+2)

Now, pick any two values for x to obtain two equations, and solve the system to find A & B.

5

Sorry I really didn’t understand that.

6

Hm, I guess that would work. I never thought about doing it like that. Here’s another way. Multiply it out:

6x[sup]2[/sup] - 5x = Ax[sup]2[/sup] + (4A + B)x + (4A + 2B + C)

This gives you three simultaneous equations, one for each power of x:

6 = A
-5 = 4A + B
0 = 4A + 2B + C