Find the partial fraction decomposition of the rational function.
5x^2 + 8/x3 + x2

- anonymous

Find the partial fraction decomposition of the rational function.
5x^2 + 8/x3 + x2

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- anonymous

Is this \(\frac{5x^2+8}{x^3+x^2}\)?

- anonymous

yes

- anonymous

got a 5x^2 + 8 = A/x^2 + B/x + 1
A(x+1) + B(x^2)
(Ax+A) + (Bx^2)
(Ax + Bx^2) + A
5x^2 + 8 = (A + B) x + (A)x^2
5x^2+x+8=(A+B) + (A)

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## More answers

- anonymous

but i think im wrong

- anonymous

the x^2 is throwing me off

- anonymous

So the problem is in how you split it up in the first step. First you've got to realize that the denominator splits into a factor that gets taken into a power. Namely, \(x^3+x^2\) becomes \(x^2(x+1)\), where \(x\) is a factor taken to the second power. The split up thus goes as follows:
\[
\begin{align*}
\frac{5x^2+8}{x^3+x^2} &= \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1}\\
5x^2+8 &= Ax(x+1) + B(x+1) + Cx^2
\end{align*}
\]

- anonymous

where did the A/x come from?

- anonymous

So the \(\frac{A}{x}\) comes from the fact that the \(x\) factor is taken to a power. As another example, if we had taken \(\frac{1}{(x+1)^2}\), the decomposition would split into \(\frac{A}{x+1} + \frac{B}{(x+1)^2}\).

- anonymous

oh ok

- anonymous

Can you solve the problem from there, then?

- anonymous

im trying it

- anonymous

shouldnt the next step be 5x^2 + 8= (Ax^2+Cx^2) + (Ax+Bx) +B?
After combining terms?

- anonymous

Exactly.

- anonymous

Ok thanks for your help!

- anonymous

Is -7/x + 8/x^2 + 12/x+1 the right answer?

- anonymous

I got a different answer from that. You can see that something is off because \(A+B\) should equal \(0\), as there is no \(x\) term on the left hand side.

- anonymous

im lost.

- anonymous

So how did you go about solving once you combined terms?

- anonymous

I'll show you

- anonymous

5x^2+8=(A+C)x^2 + (A+B)x + B
5x^2 +0x+8=(A+C) + (A+B) + B
then i put it on the matrix function on my calculator and it gave me -7, 8 and 12.

- anonymous

thats how i got Is -7/x + 8/x^2 + 12/x+1.

- anonymous

You probable plugged some numbers in wrong. I get -8, 8, and 13.

- anonymous

probably*

- anonymous

let me see if that works.

- anonymous

Thats it. I dont understand what i did arong though?

- anonymous

I had 1 0 1 5
1 1 0 1
0 1 0 8
as my matrix.

- anonymous

Row 2, column 4 should be a 0, not a one, as there's no \(x\) term.

- anonymous

Ah i see the 0x term stays zero and not 1.

- anonymous

Exactly.

- anonymous

Alright I have now. Thank you so much for helping me. I really appreciate it!

- anonymous

No problem, glad you understood it!

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