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

- anonymous

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

- jamiebookeater

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- anonymous

can you factor the denominator?

- anonymous

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

- anonymous

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

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## More answers

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yes of course

- anonymous

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?

- anonymous

Yes that's completely fine. thank you

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

one minute i am reading your work

- anonymous

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

- anonymous

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

- anonymous

ok good

- anonymous

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

- anonymous

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

- anonymous

yes

- anonymous

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

- anonymous

you can do that, but first lets think

- anonymous

Ok..

- anonymous

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

- anonymous

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

- anonymous

Sorry I do not understand why Ax^3 = 0

- anonymous

because the numerator is \(x^2\)

- anonymous

no \(x^3\) in it

- anonymous

that means \(A=0\)

- anonymous

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

- anonymous

look at the original question

- anonymous

Ok

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

ooh no scratch that, i am totally wrong, sorry

- anonymous

No worries.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

correct?

- anonymous

yeah, now lets find \(B\)

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Ok. That makes sense.

- anonymous

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

- anonymous

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

- anonymous

Ok makes sense.

- anonymous

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

- anonymous

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

- anonymous

Ok. Why does the right side equal 0 though?

- anonymous

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

- anonymous

the constant is 1

- anonymous

no

- anonymous

Sorry I mean coefficient. yes there is no constant

- anonymous

right

- anonymous

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 \(0\)

- anonymous

Makes sense now.

- anonymous

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

- anonymous

ok good

- anonymous

You made the constants equal to each other.

- anonymous

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

- anonymous

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

- anonymous

that is one way to write it, yes

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yw

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