Question

Find the partial fraction decomposition of the rational function.
9x^2 − 3x − 9
x^4 + 3x3

Answer 1

are you looking for an answer or the process?

Answer 2

both

Answer 3

well first your need to set up the equivalence \[\frac{9x^2-3x-9}{x^3(x+3)}=\]\[\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x+3}\]

Answer 4

ok

Answer 5

Thats whats confusing me

Answer 6

what is confusing me is that this interface is buggy

Answer 7

now it works...

Answer 8

yea

Answer 9

then rewrite that as
\[9x^2-3x-0=\]\[Ax^2(x+3)+Bx(x+3)\]\[+C(x+3)+Dx^3\]

Answer 10

that should be a 9 where that 0 is

Answer 11

k

Answer 12

then rewrite that as
\[9x^2-3x-9=\]\[Ax^2(x+3)+Bx(x+3)\]\[+C(x+3)+Dx^3\]

Answer 13

so are we good to this point?

Answer 14

yes

Answer 15

now we can choose x=0 and determine that C=-3

Answer 16

ok

Answer 17

and next let x=-3 to determine that D=-3

Answer 18

are you good with what I am doing?

Answer 19

yea i get it

Answer 20

now however we are going to need to choose two different values of x and get a system of two equations in two variable tosolve for A and B

Answer 21

alright

Answer 22

so I chose x=1 and got 3=A+B

Answer 23

Ok

Answer 24

next I chose x=-1 and got 3=A-B

Answer 25

adding those equations together I get 6=2A which says A=3

Answer 26

then using A=3 in either of the other two either of the previous two equations I determine that B=0

Answer 27

Now finally substituting those values for A,B,C,D we get\[\frac{9x^2-3x-3}{x^3(x+3)}=\]\[\frac{3}{x}-\frac{3}{x^3}-\frac{3}{x+3}\]

Answer 28

ok i see

Answer 29

Thats alot!

Answer 30

your welcome, glad you followed it. Long process...

Answer 31

oh yea