1. ## Partial Fraction problem

Here's the thing, I know how to do a partial fraction problem, but this particular one doesn't seem to make any sense. The problem is given to you already broken up into the fractions, and asks you to find the values of A, B, and C:

$\frac{1}{x^3 - 5x^2} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-5}$

Now here's the weirdness. If you multiply $x * x^2 * x-5$ back together, you don't get $x^3 - 5x^2$.

Because of this, I can't seem to do the problem, because when I get the system of equations for A, B, and C, I have nothing to set equal to 1! Am I just not understanding this correctly, or is the problem itself flawed?

2. Originally Posted by lysserloo
Here's the thing, I know how to do a partial fraction problem, but this particular one doesn't seem to make any sense. The problem is given to you already broken up into the fractions, and asks you to find the values of A, B, and C:

$\frac{1}{x^3 - 5x^2} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-5}$

Now here's the weirdness. If you multiply [tex]x * x^2 * x-5[tex] back together, you don't get $x^3 - 5x^2$. Mr F says: You're not meant to. Go back and review your class notes on what to do when there is a repeated factot (the repeated factor in your case is x).

Because of this, I can't seem to do the problem, because when I get the system of equations for A, B, and C, I have nothing to set equal to 1! Am I just not understanding this correctly, or is the problem itself flawed?
..

3. OH PFFF. Wow. I didn't even realize that. Thank you very much!

EDIT: Even using $x^2(x-5)$...this doesn't make sense once I have to assign a value to the system of equations. After multiplying through I get:

$Ax^3 + (-5Ax + C)x^2 + Bx -5B$

So the system of equations I get to solve are:

A = 0 , -5A + C = 0, B = 0, -5B = 1

I know that -5B = 1 is correct, and so B = -1/5

But the other equations that equal 0 don't make any sense. I know A is not equal to 0, and B is OBVIOUSLY not equal to 0!

What am I doing wrong?

4. Hi Lisserloo,

if you multiply by $\frac{x}{x}$ everything works out!

5. What exactly am I multiplying by $\frac{x}{x}$?

6. Originally Posted by lysserloo
OH PFFF. Wow. I didn't even realize that. Thank you very much!

EDIT: Even using $x^2(x-5)$...this doesn't make sense once I have to assign a value to the system of equations. After multiplying through I get:

$Ax^3 + (-5Ax + C)x^2 + Bx -5B$

So the system of equations I get to solve are:

A = 0 , -5A + C = 0, B = 0, -5B = 1

I know that -5B = 1 is correct, and so B = -1/5

But the other equations that equal 0 don't make any sense. I know A is not equal to 0, and B is OBVIOUSLY not equal to 0!

What am I doing wrong?
$\frac{1}{x^3 - 5x^2} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-5} = \frac{Ax(x-5)}{x^3-5x^2} + \frac{B(x-5)}{x^3 - 5x^2} + \frac{Cx^2}{x^3-5x^2}$

7. Just in case:

\begin{aligned}
\frac{1}{x^{3}-5x^{2}}&=\frac{1}{x^{2}(x-5)} \\
& =\frac{1}{25}\cdot \frac{x^{2}-(x+5)(x-5)}{x^{2}(x-5)} \\
& =\frac{1}{25}\left( \frac{1}{x-5}-\frac{1}{x}-\frac{5}{x^{2}} \right).
\end{aligned}

8. Originally Posted by lysserloo
What exactly am I multiplying by $\frac{x}{x}$?
$\frac{x}{x}\ \frac{1}{x^3-5x^2}$

mr fantastic has recommended a clearer way, though.