Question

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

Answer 1

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

Answer 2

yes

Answer 3

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)

Answer 4

but i think im wrong

Answer 5

the x^2 is throwing me off

Answer 6

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:
\[
\begin{align*}
\frac{5x^2+8}{x^3+x^2} &= \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1}\\
5x^2+8 &= Ax(x+1) + B(x+1) + Cx^2
\end{align*}
\]

Answer 7

where did the A/x come from?

Answer 8

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 \(\frac{A}{x+1} + \frac{B}{(x+1)^2}\).

Answer 9

oh ok

Answer 10

Can you solve the problem from there, then?

Answer 11

im trying it

Answer 12

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

Answer 13

Exactly.

Answer 14

Ok thanks for your help!

Answer 15

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

Answer 16

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.

Answer 17

im lost.

Answer 18

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

Answer 19

I'll show you

Answer 20

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.

Answer 21

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

Answer 22

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

Answer 23

probably*

Answer 24

let me see if that works.

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

Exactly.

Answer 30

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

Answer 31

No problem, glad you understood it!