Question

Partial Fraction decomposition:
(x^2)/(x^4-2x^2-8)

Answer 1

can you factor the denominator?

Answer 2

Yes i factored denominator: (x^2+2)(x-2)(x+2)

Answer 3

ok then we have \[\frac{x^2}{(x^2+2)(x-2)(x+2)}\] \[=\frac{Ax+B}{x^2+2}+\frac{C}{x-2}+\frac{D}{x-2}\] as a fist step

Answer 4

actually i think in this case \(A=0\) but it doesn't need to be

Answer 5

would (A/x-2) + (B/x+2) + ((Cx+D)/(x^2+2)) be correct as well?

Answer 6

now add \[Ax+B(x+2)(x-2)+C(x^2+2)(x-2)+D(x^2+2)(x+2)=x^2\]

Answer 7

yes of course

Answer 8

it doesn't matter since addition is commutative
but since i spent all that time writing it in latex, would you mind using it so i don't have to start over?

Answer 9

Yes that's completely fine. thank you

Answer 10

so now there are a couple of ways to proceed
the easiest way is probably to replace \(x\) by \(2\) and see what you get

Answer 11

the first two terms drop out, you are left with \[D(2^2+2)(2+2)=2^2\] and you can solve that for \(D\)

Answer 12

that is really the basic gimmick
is it clear what i did? you get \[D=\frac{1}{6}\] pretty much in your head

Answer 13

one minute i am reading your work

Answer 14

ok take your time
i am going to get a snack
if it is not clear, ask

Answer 15

I understand what you did. I worked the problem and got D=1/6

Answer 16

ok good

Answer 17

now we can use the same gimmick only replace \(x\) yb \(-2\)

Answer 18

Ok, so that cancels out the first and last term. I can now solve for C. C=-1/6?

Answer 19

yes

Answer 20

and now I plugin C and D into the first equation to solve for A & B?

Answer 21

you can do that, but first lets think

Answer 22

Ok..

Answer 23

the first part is \[(Ax+B)(x+2)(x-2)\]

Answer 24

if you multiply that out mentally you get an \(Ax^3\) term
there is not \(x^3\) so \(A=0\)

Answer 25

Sorry I do not understand why Ax^3 = 0

Answer 26

because the numerator is \(x^2\)

Answer 27

no \(x^3\) in it

Answer 28

that means \(A=0\)

Answer 29

So the exponent in the denominator cannot be greater than x^2?

Answer 30

look at the original question

Answer 31

Ok

Answer 32

or even just this part \[(Ax+B)(x+2)(x-2)+C(x^2+2)(x-2)+D(x^2+2)(x+2)=x^2\]

Answer 33

on the right hand side of the equal sign, there is only a little lonely \(x^2\)
no \(x^3\) at all

Answer 34

on the left side there will be, when you multiply out, one \(x^3\) term
it will be \(Ax^3\)

Answer 35

So A must equal 0 in order for the equation to be true

Answer 36

ooh no scratch that, i am totally wrong, sorry

Answer 37

No worries.

Answer 38

there are other \(x^3\) terms there i forgot about them

Answer 39

so yeah, substitute back what you know to solve for \(A\) and\(B\)

Answer 40

you will find that \(A=0\) in any case

Answer 41

\[x^2 = Ax+B(x+2)(x-2) - (1/6)(x^2+2)(x-2)+(1/6)(x^2+2)(x+2)\]

Answer 42

correct?

Answer 43

yeah, now lets find \(B\)

Answer 44

without doing a raft of solving a system of equations, which is the other long and tedious method of doing this

Answer 45

\[(Ax+B)(x+2)(x-2)-\frac{1}{6}(x^2+2)(x-2)+\frac{1}{6}(x^2+2)(x+2)=x^2\]

Answer 46

what will the constant be? the term without any \(x\) or any \(x^2\)?

Answer 47

if it is not clear what i mean, let me know

Answer 48

Ok i will work on the problem and let you know.

Answer 49

Ok, I am confused. I am stuck on the part where you said "what will the constant be? the term without any x"

Answer 50

from here \[-\frac{1}{6}(x^2+2)(x-2)\] when you mutiply out, you will get a number without an \(x\) namely \[-\frac{1}{6}\times 2\times (-2)=\frac{2}{3}\]

Answer 51

from here \[\frac{1}{6}(x^2+2)(x+2)\] you will also get \(\frac{2}{3}\)

Answer 52

Ok. That makes sense.

Answer 53

from here \[(Ax+B)(x+2)(x-2)\] then number will be \(4B\)

Answer 54

]scratch that it will be \[-4B\]

Answer 55

Ok makes sense.

Answer 56

on the right hand side there is no constant at all,

Answer 57

that means \[\frac{2}{3}+\frac{2}{3}-4B=0\]or \[4B=\frac{4}{3}\] making \[B=\frac{1}{3}\]

Answer 58

Ok. Why does the right side equal 0 though?

Answer 59

ok i was mistaken before when i was trying to explain why \(A=0\) but now i am not
the original numerator is just \(x^2\)
it has no constant in it i

Answer 60

the constant is 1

Answer 61

no

Answer 62

Sorry I mean coefficient. yes there is no constant

Answer 63

right

Answer 64

you got \[(Ax+B)(x+2)(x-2)-\frac{1}{6}(x^2+2)(x-2)+\frac{1}{6}(x^2+2)(x+2)=x^2\] the contant on the left is \[\frac{4}{3}-4B\] and on the right it is \(0\)

Answer 65

Makes sense now.

Answer 66

that makes \(B=\frac{1}{3}\)

Answer 67

ok good

Answer 68

You made the constants equal to each other.

Answer 69

So I get the answer:
[(1/3)/(x^2+2)] + [(-1/6)/(x+2)] + [(1/6)/(x-2)]
?

Answer 70

yes
sometimes you can get away without doing that, but not in this case because of the \(x^2+2\) which is never zero so you can't use the previous trick of substitution

Answer 71

that is one way to write it, yes

Answer 72

Another way to write it would be [1/3(x^2+2)] - [1/6(x+2)] + [1/6(x-2)]
?

Answer 73

So would my answer that I put be correct as well?

Answer 74

normally you would write \[\frac{1}{3(x^2+2)}-\frac{1}{6(x+2)}+\frac{1}{6(x-2)}\]

Answer 75

it is the same
only in this topic do you see compound fractions allowed as an answer, but yes, both are the same

Answer 76

Ok that makes sense. Thank you so much for you time I appreciate it. Good luck with your studies

Answer 77

yw