Partial Fraction Decomposition

[CornMuffin]

Homework Statement

To find the decomposition of a polynomial with a repeated factor in the denominator, you should separate them into (x+a)^1 + ... + (x+a)^n. But, my question is why?

For example, why should you decompose it in the following way:
[tex] \frac{x+2}{(x+1)(x+3)^2} = \frac{A}{x+1} + \frac{B}{x+3} + \frac{C}{(x+3)^2}[/tex]

why, for example, isn't it decomposed into either of the following ways:
[tex] \frac{x+2}{(x+1)(x+3)^2} = \frac{A}{x+1} + \frac{Bx+C}{(x+3)^2}[/tex]
OR
[tex] \frac{x+2}{(x+1)(x+3)^2} = \frac{A}{x+1} + \frac{B}{x+3}[/tex]

 

[Alexander2357]

[tex]\frac{x+2}{(x+1)(x+3)^{2}}\equiv \frac{A}{(x+1)}+\frac{Bx+D}{(x+3)^{2}}[/tex]
[tex]\equiv \frac{A}{(x+1)}+\frac{B(x+3)-3B+D}{(x+3)^{2}}[/tex]
[tex]\equiv \frac{A}{(x+1)}+\frac{B}{(x+3)}+\frac{-3B+D}{(x+3)^{2}}[/tex]
[tex]\equiv \frac{A}{(x+1)}+\frac{B}{(x+3)}+\frac{C}{(x+3)^{2}}[/tex]

 

[CornMuffin]

[tex]\frac{x+2}{(x+1)(x+3)^{2}}\equiv \frac{A}{(x+1)}+\frac{Bx+D}{(x+3)^{2}}[/tex]
[tex]\equiv \frac{A}{(x+1)}+\frac{B(x+3)-3B+D}{(x+3)^{2}}[/tex]
[tex]\equiv \frac{A}{(x+1)}+\frac{B}{(x+3)}+\frac{-3B+D}{(x+3)^{2}}[/tex]
[tex]\equiv \frac{A}{(x+1)}+\frac{B}{(x+3)}+\frac{C}{(x+3)^{2}}[/tex]

Ok thank you so much, I understand.

But, you made a small mistake. I fixed it above

 

[Mark44]

[tex]\frac{x+2}{(x+1)(x+3)^{2}}\equiv \frac{A}{(x+1)}+\frac{Bx+D}{(x+3)^{2}}[/tex]

You don't need to do this for repeated linear factors, so most of that work can be skipped. You can start directly from this and skip most of the work below:
$$\frac{x+2}{(x+1)(x+3)^{2}}\equiv \frac{A}{x+1}+\frac{B}{x+3} + \frac{C}{(x + 3)^2}$$
A numerator of Bx + D would be used for an irreducible quadratic factor such as x² + 1.

[tex]\equiv \frac{A}{(x+1)}+\frac{B(x-1)+B+D}{(x+3)^{2}}[/tex]
[tex]\equiv \frac{A}{(x+1)}+\frac{B}{(x+3)}+\frac{B+D}{(x+3)^{2}}[/tex]
[tex]\equiv \frac{A}{(x+1)}+\frac{B}{(x+3)}+\frac{C}{(x+3)^{2}}[/tex]

 

[Alexander2357]

Ok thank you so much, I understand.

But, you made a small mistake. I fixed it above

You are welcome.

 

[Alexander2357]

You don't need to do this for repeated linear factors, so most of that work can be skipped.

Yes, definitely, it can all be skipped but the question is specifically asking for why it works.

 

[SammyS]

Yes, definitely, it can all be skipped but the question is specifically asking for why it works.

Either use long division on ##\displaystyle \ \frac{Bx+C}{x+3}\,, \ ## and then multiply the result by
##\displaystyle \ \frac{1}{x+3}\ .##

You could use the following rather than long division. Mark44 seems to be a big fan of it.

Rewrite ##\displaystyle \ \frac{x+K}{x+3}\ ## as ##\displaystyle \ \frac{x+3+K-3}{x+3}\ ## which splits into ##\displaystyle \ \frac{x+3}{x+3}+\frac{C-3}{x+3}\,, \ ## the first term being 1 .


(Mess around with the B & C as you please.)​

Alternatively, use a common denominator to combine ##\ \displaystyle \frac{B}{x+3} + \frac{C}{(x+3)^2}\ ## into one rational expression.

 

[vela]

For example, why should you decompose it in the following way:
[tex] \frac{x+2}{(x+1)(x+3)^2} = \frac{A}{x+1} + \frac{B}{x+3} + \frac{C}{(x+3)^2}[/tex]

why, for example, isn't it decomposed into either of the following ways:
[tex] \frac{x+2}{(x+1)(x+3)^2} = \frac{A}{x+1} + \frac{Bx+C}{(x+3)^2}[/tex]
OR
[tex] \frac{x+2}{(x+1)(x+3)^2} = \frac{A}{x+1} + \frac{B}{x+3}[/tex]

For the latter case, if we multiply by (x+3), the lefthand side becomes
$$\frac{x+2}{(x+1)(x+3)}$$ while the righthand side becomes
$$A \frac{x+3}{x+1} + B.$$ The expression on the lefthand side has a vertical asymptote at ##x=-3##, but the one from the righthand side doesn't, which means the two sides can't be equal for all ##x##. This implies that the original decomposition isn't valid.