Partial Fractions - 3 Unknowns

In summary: That must be exactly the same as x- 1 so Ax^2 must be 0x^2, 4Ax must be x, and 4A- 1 must be -1. That gives A= -2. Now multiply out the right hand side with A= -2: -2x^2- 8x- 8+ Bx^2+ Bx+ Cx+ C= 0x^2+ x- 1= x- 1. So we must have -2x^2+ Bx^2= 0x^2 so B= 2. -8x+ Bx= -7x so we must have -
  • #1
BOAS
553
19
Hello,

i've come across a partial fractions problem that I don't know how to solve - Usually, the denominator of the fraction I need to split up into two separate fractions is a quadratic, but in this instance it's a cubic.

Specifically, the problem I'm having is that two of the factors to this cubic are the same, so when I set x equal to values in order to find the unknowns, everything disappears!

Homework Statement



Write [itex]\frac{(x - 1)}{(x + 1)(x + 2)^{2}}[/itex] as the sum of 3 partial fractions.

Homework Equations


The Attempt at a Solution



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

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

[itex](x - 1) \equiv A(x +2)(x + 2) + B(x+1)(x + 2) + C(x + 1)(x + 2)[/itex]

If I set x = -1, then A = -2

But now if I try to make the other brackets equal to zero, A, B and C disappear...

I'm probably missing something obvious (or shouldn't be approaching this in the exact same way I would a simpler one), but I can't see the solution.

Thanks for any help you can give
 
Last edited:
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  • #2
In this case, the correct form of the partial fraction decomposition is ##\frac{x-1}{(x+1)(x+2)^2}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}##. You can calculate the correct decomposition in Wolfram Alpha by entering the command Apart[(x-1)/((x+1)(x+2)^2)]. I hope this helps.
 
  • #3
hilbert2 said:
In this case, the correct form of the partial fraction decomposition is ##\frac{x-1}{(x+1)(x+2)^2}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}##. You can calculate the correct decomposition in Wolfram Alpha by entering the command Apart[(x-1)/((x+1)(x+2)^2)]. I hope this helps.

I'm not sure I understand why that's the case, but I can see how it helps me solve the question.
 
  • #4
BOAS said:

Homework Statement



Write [itex]\frac{(x - 1)}{(x + 2)(x + 2)^{2}}[/itex] as the sum of 3 partial fractions.

hilbert2 said:
In this case, the correct form of the partial fraction decomposition is ##\frac{x-1}{(x+1)(x+2)^2}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}##. You can calculate the correct decomposition in Wolfram Alpha by entering the command Apart[(x-1)/((x+1)(x+2)^2)]. I hope this helps.

BOAS said:
I'm not sure I understand why that's the case, but I can see how it helps me solve the question.

It's not clear what partial fraction decomposition you are trying to write.

BOAS starts with:

[itex]\frac{(x - 1)}{(x + 2)(x + 2)^{2}}[/itex]

Which hilbert2 interprets as:

[itex]\frac{x-1}{(x+1)(x+2)^2}[/itex]

and then, the partial fraction decomposition becomes (also, according to hilbert2):

[itex]\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}[/itex]

Are you guys solving the same problem?
 
  • #5
SteamKing said:
It's not clear what partial fraction decomposition you are trying to write.

BOAS starts with:

[itex]\frac{(x - 1)}{(x + 2)(x + 2)^{2}}[/itex]

Which hilbert2 interprets as:

[itex]\frac{x-1}{(x+1)(x+2)^2}[/itex]

and then, the partial fraction decomposition becomes (also, according to hilbert2):

[itex]\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}[/itex]

Are you guys solving the same problem?

Thank you for catching that.

In the 'problem statement' section, I did make a typo, but the correct version is used through my working and Hilbert references the correct version.

I'll edit the OP.

EDIT 2 Ack, there are more inconsistencies in my post, further edits required.
 
  • #6
Seems like every post in this thread has a typo or two.

BOAS said:
Hello,

i've come across a partial fractions problem that I don't know how to solve - Usually, the denominator of the fraction I need to split up into two separate fractions is a quadratic, but in this instance it's a cubic.

Specifically, the problem I'm having is that two of the factors to this cubic are the same, so when I set x equal to values in order to find the unknowns, everything disappears!

Homework Statement



Write [itex]\frac{(x - 1)}{\color{red}{(x + 2)}(x + 2)^{2}}[/itex] as the sum of 3 partial fractions.

Homework Equations





The Attempt at a Solution



[itex]\frac{(x - 1)}{\color{red}{(x + 1)}(x + 2)^{2}} = \frac{A}{(x - 1)} + \frac{B}{(x + 2)} + \frac{C}{(x + 2)}[/itex]

I assume the second red factor is correct. In that case the right side should be$$
\frac A {x+1} + \frac B {x+2} + \frac C {(x+2)^2}$$.
 
  • #7
LCKurtz said:
Seems like every post in this thread has a typo or two.
I assume the second red factor is correct. In that case the right side should be$$
\frac A {x+1} + \frac B {x+2} + \frac C {(x+2)^2}$$.

Ok, I think I have corrected all the typos in my OP.

I've also answered my own problem;

[itex]\frac{(x - 1)}{(x + 1)(x + 2)^{2}} = \frac{A}{(x + 1)} + \frac{Bx + c}{(x + 2)^{2}}[/itex]

Letting [itex]C = 2B + D[/itex] Accounts for that extra factor that was confusing me.

Thanks for the help - I made the same sign error early on in my written working and it cropped up all over the place.
 
  • #8
There are three unknowns, A, B, and C so you need three equations. Since the given equation is to be true for all x, put in any three values of x you please to get three equations.

You have
[tex]\frac{x- 1}{(x+ 1)(x+ 2)^2}= \frac{A}{x+ 1}+ \frac{Bx+ C}{(x+ 2)^2}[/tex]
Multiply both sides by the denominator to get [itex]x- 1= A(x+ 2)^2+ (Bx+ C)(x+ 1)[/itex]

Yes, setting x= -1 and -2 will give very simple equations. If x= -1, [itex]-2= A(1)^2+ (B(-1)+ C)(-1+ 1)= A[/itex]. Setting x= -2, [itex]-3= A(-2+ 2)^3+ (B(-2)+ C)(-2+ 1)= 2B- C[/itex].

So we have A= -2 and 2B- C= -3. Set x equal to what ever third number you like to get a third equation. I happen to thing "0" is a very easy number: if x= 0, -1= A(0+ 2)^2+ (B(0)+ C)(0+ 1)= 4A+ C.

So we have the three equations, A= -2, 2B- C= -3, and 4A+ C= -1. Since A= -2, that third equation is -8+ C= 1 so C= 9. Then 2B- C= 2B- 9= -3.

Another way to do problems like this, without setting x equal to any numbers is use the fact that "If two polynomials have the same value for all x, then they must be exactly the same- they must have the same coefficients. We have [itex]x- 1= A(x+ 2)^2+ (Bx+ C)(x+ 1)[/itex]. Multiplying out the right side, [tex]0x^3+ x- 1= A(x^2+ 4x+ 4)+ Bx(x+ 1)+ C(x+ 1)= Ax^2+ 4Ax+ 4A+ Bx^2+ Bx+ Cx+ C= (A+ B)x^2+ (B+ C)x+ (A+C)[/tex] for all x so we must have A+ B= 0, B+ C= 1, and 4A+ C= -1. Solve those three equations for A, B, C.
 

FAQ: Partial Fractions - 3 Unknowns

1. What are partial fractions with 3 unknowns?

Partial fractions with 3 unknowns are a mathematical technique used to break down a complex rational expression into simpler terms. This is accomplished by expressing the rational expression as a sum of simpler fractions with unknown coefficients.

2. When are partial fractions with 3 unknowns used?

Partial fractions with 3 unknowns are commonly used in integration and differential equations. They are also useful in solving linear equations and simplifying algebraic expressions.

3. How do you find the unknown coefficients in a partial fraction with 3 unknowns?

To find the unknown coefficients in a partial fraction with 3 unknowns, you must first equate the original rational expression to the sum of the simpler fractions. Then, you can use algebraic manipulation and substitution to solve for the unknown coefficients.

4. Can partial fractions with 3 unknowns be solved by hand?

Yes, partial fractions with 3 unknowns can be solved by hand using algebraic techniques such as equating coefficients, substitution, and manipulation. However, for more complex expressions, it may be easier to use a computer or calculator to solve for the unknown coefficients.

5. What is the purpose of using partial fractions with 3 unknowns?

The purpose of using partial fractions with 3 unknowns is to simplify complex rational expressions into simpler terms. This makes it easier to perform mathematical operations, such as integration or solving equations, and can provide a better understanding of the relationships between the variables in the expression.

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