# Partial Fraction Decomposition Problem

1. Dec 10, 2007

### ko_kidd

$$\frac{7}{3s^{2}(3s+1)}$$

Can this be decomposed, and how?

2. Dec 10, 2007

### symbolipoint

I'd say, yes; it can be decomposed; without my first relearning the method and trying to decompose to partial fractions. Your denominators might be $$$3s^2$$$ and $$$3s + 1$$$

3. Dec 10, 2007

### ko_kidd

Yep. I understood that much, but I'm confused about what I can do from there because I end up with:

$$\frac{A}{3s^{2}}$$ + $$\frac{B}{3s+1}$$

which would equal out to

$$\frac{A}{3s^{2}}$$ + $$\frac{B}{3s+1}$$ = $$\frac{7}{3s^{2}(3s+1)}$$

then

7 = $$A(3s+1)$$ + $$3Bs^{2}$$

--which is where I'm stuck, because eliminating "s" leaves me with A = 7, which doesn't make sense if I want to separate the fraction.

4. Dec 10, 2007

### Big-T

It is important to remember that if one of the factors are squared, then you need to fractions for that expression, i.e. for (7/3)//(s^2(3s+1)), you need to decompose it into a/s AND b/s^2, in addition to c/(3s+1).

5. Dec 10, 2007

### gao xiong

$$\frac{7}{3s^{2}(3s+1)}=\frac{As+B}{3s^{2}}+\frac{C}{3s+1}$$
A=-21
B=7
C=21

Last edited: Dec 10, 2007
6. Dec 10, 2007

### symbolipoint

You may need to fill in some missing steps, but this seemed to work:
$$$\begin{array}{l} \frac{a}{{3s^2 }} + \frac{b}{{3s + 1}} = performSteps = \frac{{(3s + 1)a + 3s^2 b}}{{3s^2 (3s + 1)}} \\ fromOriginalDeno\min ator,\;0s + 7 = 3s^2 b + 3sa + a \\ Answer:\;\;a = 7\quad b = \frac{{ - 7}}{s} \\ \end{array}$$$

7. Dec 10, 2007

### ko_kidd

Hmm, I'll have to try to this in a second.

--wow the method gao xiong did works but I don't recall seeing this in my initial searches through the textbook or online.

Nice method, thanks.

Big-T, I don't understand where the a/s would come from if the highest polynomial is 9s^3 (or the highest term is cubic) when you combine the denominator.

Last edited: Dec 10, 2007
8. Dec 10, 2007

### symbolipoint

gao_xiong's form makes the best sense, and he obtained constants as answers. He decided that $$$3s^2$$$
is an irreducible quadratic expression and gave a suitable linear binomial expression with it.

9. Dec 11, 2007

### ko_kidd

another problem that's confusing me

I have one more problem.

Would this:$$\frac{87}{(x)(x^{2}+13x+38)}$$

simplify to something like

$$\frac{Ax+B}{x^{2}+13x+38} + \frac{C}{x}$$ = $$\frac{87}{(x)(x^{2}+13x+38)}$$

10. Dec 11, 2007

### sutupidmath

well first look if $$\{x^{2}+13x+38}$$ has any roots over reals so the problem could get more simplified, since you could express that to some form of (x+-a)(x+-b) where a,b are the real roots.
then it would be of the form
$$\frac{A}{x+-a} + \frac{B}{x+-b}+\frac{C}{x}$$

Last edited: Dec 11, 2007
11. Dec 11, 2007

### HallsofIvy

With that "s2", you are going to need both 1/s and 1/s2.
$$\frac{7}{3s^2(3s+1)}= \frac{A}{s}+ \frac{B}{s^2}+ \frac{C}{3s+1}$$
Multiplying through by the common denominator, $7= As(3s+1)+ B(3s+1)+ Cs^2$. Taking s= 0, 7= B. Taking s= -1/3, 7= C/9 so C= 63. Finally, taking s= 1, 7= 4A+ 4B+ C= 4A+ 28+ 63. 4A= 7- 91= -84, A= -21.

12. Dec 11, 2007

### ko_kidd

I don't doubt what you're saying, but how do [you] rationalize

a 1/s and 1/s^2 from one 1/s^2. I've never seen that before.

Last edited: Dec 11, 2007
13. Dec 11, 2007

### BSMSMSTMSPHD

You lost the 7 in the original numerator. The correct values are A = -3, B = 1, C = 9.