Hi, I'm working on the decomposition of this partial fraction and I solved for two of the constant. I cant seem to solve for the constant "A". Can someone please help me with this one?

Thanks,

Coop

2. Originally Posted by coopsterdude
Hi, I'm working on the decomposition of this partial fraction and I solved for two of the constant. I cant seem to solve for the constant "A". Can someone please help me with this one?

Thanks,

Coop
I'd do it this way (my preferred method):
$4x^2 + 2x - 1 = A(x^2 + x) + B(x + 1) + Cx^2$

$4x^2 + 2x - 1 = (A + C)x^2 + (A + B)x + B$

So matching coefficients we get
$A + C = 4$

$A + B = 2$

$-1 = B$
which you can solve easily.

-Dan

3. Hello, Coop!

I've never seen a layout like yours.
. . [I wonder where your teacher (or book) learned that approach.]
I've seen topsquark's method many times.

I'll show you the method I was taught . . .

$\frac{4x^2+2x-1}{x^2(x+1)} \;=\;\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1}$

Multiply through by the LCD: . $4x^2 + 2x-1 \;=\;x(x+1)A + (x+1)B + x^2C$

Now we select some "good" values for $x.$

Let $x = 0:\;\;\text{-}1 \:=\:0\!\cdot\!A + 1\!\cdot\!B + 0\!\cdot\!\quad\Rightarrow\quad\boxed{B \,=\,\text{-}1}$

Let $x = \text{-}1:\;\;1 \:=\:0\!\cdot\!A + 0\!\cdot\!B + 1\!\cdot\!C\quad\Rightarrow\quad\boxed{C \,=\,1}$

Let $x = 1:\;\;5 \:=\:2A + 2B + C$
. . Since $B = \text{-}1,\:C = 1$, we have: . $5 \:=\:2A - 2 + 1\quad\Rightarrow\quad\boxed{A \,=\, 3}$

Therefore: . $\frac{4x^2+2x-1}{x^2(x+1)} \;=\;\frac{3}{x} - \frac{1}{x^2} + \frac{1}{x+1}$

4. Well, I think he did the same thing. He just multiplied every term by the denominator and wrote it, instead of writing the result directly.