Question

Is this partial fraction decomposition correct?

Answer 1

\[\frac{ 1 }{ x^3 +2x^2+x} = \frac{ A }{ x }+\frac{ B }{ x+1 }+\frac{ C }{ (x+1)^2 }\]

Answer 2

\(\color{#000000}{ \displaystyle \frac{1}{x^3+2x^2+x} =\frac{1}{x(x+1)(x+1)} }\)

Answer 3

Your factors are all linear, so the generic terms are all constants. Yes, your answer is correct.

Answer 4

i got A=1 but how do i find B and C?

Answer 5

If you had something like \(x^2+2\), then this is not a linear times linear, and in that case, you would need to write \(\color{#000000}{ \displaystyle Ax+B }\) in numerator.

Answer 6

Ok, now, how to find A, B, C.

Answer 7

\(\color{#000000}{ \displaystyle \frac{1}{x^3+2x^2+x} =\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2} }\)

multiply all through by the big denominator (by \(\color{#000000}{ x^3+2x^2+x }\)), and then show me what you get.

Answer 8

\[A(x+1)(x+1)^2+B(x)(x+1)^2+C(x)(x+1)=1\]

Answer 9

you are off by a factor of (x+1). it is just
\(\color{#000000}{ \displaystyle 1=a(x+1)^2+Bx(x+1)+Cx }\)

Answer 10

Now, this decomposition above is said to be true for all x, so you just need to plug in some values of x, to find A, B, and C.

Answer 11

why is it that instead of what i wrote?

Answer 12

\(\color{#000000}{ \displaystyle \color{#000000}{ \displaystyle \frac{1}{x^3+2x^2+x} =\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2} } }\)

you are multiplying each term by \(\color{#000000}{ \displaystyle x^3+2x^2+x }\),
and this is the same as multiplying times \(\color{#000000}{ x (x+1)^2 }\).

So, you are going to have:
\(\color{#000000}{ \displaystyle \color{#000000}{ \displaystyle \frac{1}{x^3+2x^2+x}\times x (x+1)^2=1 } }\)

\(\color{#000000}{ \displaystyle \color{#000000}{ \displaystyle \frac{A}{x}\times x (x+1)^2=(x+1)^2 } }\)

\(\color{#000000}{ \displaystyle \color{#000000}{ \displaystyle \frac{B}{x+1}\times x (x+1)^2=x(x+1) } }\)

\(\color{#000000}{ \displaystyle \color{#000000}{ \displaystyle \frac{C}{(x+1)^2}\times x (x+1)^2=x } }\)

Answer 13

So, therefore, you get,

\(\color{#000000}{ \displaystyle 1=A(x+1)^2+Bx(x+1)+Cx }\)

Answer 14

See?

Answer 15

oh, I forgot to include A, B, C in those products, but I hope you get the point anyway:)

Answer 16

oh ok

Answer 17

Yes, so you have:
\(\color{#000000}{ \displaystyle 1=A(x+1)^2+Bx(x+1)+Cx }\)

Answer 18

Now, plug in different x-values to solve for A, B, C.

Answer 19

I would advise
x=0 to solve A.
x=-1 to solve for C.
x=2, and you can solve for B once you know A and C.

Answer 20

A=1, B=1/6, C=-1

Answer 21

\(\color{#000000}{ \displaystyle 1=A(x+1)^2+Bx(x+1)+Cx }\)

\(\color{#000000}{ \displaystyle 1=A(0+1)^2+B(0)(0+1)+C(0) }\)
\(\color{#000000}{ \displaystyle 1=A }\)

\(\color{#000000}{ \displaystyle 1=(x+1)^2+Bx(x+1)+Cx }\)
\(\color{#000000}{ \displaystyle 1=(-1+1)^2+B(-1)(-1+1)+C(-1) }\)
\(\color{#000000}{ \displaystyle -1=C }\)

\(\color{#000000}{ \displaystyle 1=(x+1)^2+Bx(x+1)-x }\)
\(\color{#000000}{ \displaystyle 1=(-2+1)^2+B(-2)(-2+1)+2 }\)
\(\color{#000000}{ \displaystyle 1=(-1)^2+B(-2)(-1)+2 }\)
\(\color{#000000}{ \displaystyle 1=1+2B+2 }\)
\(\color{#000000}{ \displaystyle 0=2+2B }\)
\(\color{#000000}{ \displaystyle B=-1 }\)

Answer 22

So, therefore,

A=1
B=-1
C=-1

Answer 23

\(\color{#000000}{ \displaystyle \frac{1}{x^3+2x^2+x}=\frac{A}{x}+\frac{B}{x(x+1)}+\frac{A}{(x+1)^2} }\)

\(\color{#000000}{ \displaystyle \frac{1}{x^3+2x^2+x}=\frac{1}{x}-\frac{1}{x(x+1)}-\frac{1}{(x+1)^2} }\)

Answer 24

got it :) thanks. can you check another one?

Answer 25

Sure, but you can always use wolfram to check answers too, if you like.

Answer 26

I wouldn't mind tho', but in case no one is online to check, or if you want a faster reply ...

Answer 27

ok thanks,
is there supposed to be an x in front of (x+1) in B/x(x+1) ?

Answer 28

In the denominator, yes.

Answer 29

but you didnt put the x when you wrote it out the first time

Answer 30

I think I know what you confused.

Answer 31

When I multiply by the big denominator I will get the following:
A(x+1)²+Bx(x+1)+Cx (with x in front of B, after multiplication)

The initial decomposition, tho, is:
\(\color{#000000}{ \displaystyle \frac{1}{x^3+2x^2+x} =\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2} }\)

Answer 32

right, so i thought the answer would be
\[\frac{ 1 }{ x^3+2x^2+x }=\frac{ 1 }{ x } - \frac{ 1}{ x+1 }-\frac{ 1 }{ (x+1)^2 }\]

Answer 33

i thought i just had to plug in A, B, C

Answer 34

Oh, my bad, I see!

Answer 35

Yes, I didn't remove the x when I was doing the latex.
What you wrote is correct;)

Answer 36

oh ok :) thank you!
can i show you another one, i checked my answer on wolfram but i'm unsure of the equation to find a, b, and c

Answer 37

\[\frac{ 1 }{ x^5-x^2 }=\frac{ 1 }{ x^2(x^3-1)}=\frac{ 1 }{ x^2(x+1)(x^2-x+1) }\]
\[\frac{ Ax+B }{ x^2 }+\frac{ C }{ x+1 }+\frac{ Dx+E }{ x^2-x+1 }\]
\[Ax+B(x+1)(x^2-x+1)+C(x^2-x+1)(x^2)+Dx+E(x+1)(x^2)=1\]

Answer 38

is that right?

Answer 39

The decomposition is incorrect.

Answer 40

:(

Answer 41

x is a linear term, and x^2 is a product of two linear term (x and x).
That means that you have a generic constant below.

Answer 42

is it A/x ?

Answer 43

If you had a denominator of \(x^2+3\) or other square polynomial that can't be expressed as a product of two linears, then you would have \((Ax+D)/(x^2+2)\).

However in this case, since \(x\) is a linear, and \(x^2\) is a product of two linears, therefore you get \(A/x^2\).

Answer 44

\[\frac{ A }{ x^2} + \frac{ B }{ x+1 }+\frac{ Cx+D }{ x^2-x+1 }\]

Answer 45

Yes, precisely.

Answer 46

Then you multiply by the original denominator, to solve for A, B, C, and D.

Answer 47

\[A(x+1)(x^2-x+1)+B(x^2)(x^2-x+1)+Cx+E(x^2)(x+1)=1\]

Answer 48

Check your C.
And I suppose that E was D (that one is correct, just like the rest, besides C).

Answer 49

You are multiplying everything times \(\color{#000000}{ \displaystyle x^2(x+1)(x^2-x+1) }\).
\(\color{#000000}{ \displaystyle \frac{Cx}{x^2-x+1}\times x^2(x+1)(x^2-x+1) =Cx(x+1) }\)

Answer 50

so is that \[(C(x)+D)(x^2)(x+1)\]

Answer 51

no wait...

Answer 52

good that you noticed..

Answer 53

why don't you include the D?

Answer 54

Really, \(C+Dx\), not \(Cx+D\)!

Answer 55

If you are factoring like this, make sure that your terms match.
What you wrote originally would result in \(x^3\).

Answer 56

oh! i'll be back in like 15min, is that ok?

Answer 57

I am going to be offline just about now or in 5 minutes.
It's getting late in my location.

Answer 58

ok no problem. thank you though :)

Answer 59

You can tag me 3 times, and I will post the material if I have time, but I might now...

good luck!

Answer 60

@SolomonZelman
im confused, wouldn't you multiply
\[\frac{ (Cx+D)(x^2-x+1)(x^2)(x+1)}{ x^2-x+1 }=(Cx+D)(x^2)(x+1) \]

Answer 61

Oh, yeah you are correct, I made a mistake.
(Sorry for this delay, I was offline.)

Answer 62

thank you :)