Partial Fractions: Exponent on Denominator Explained

[xeon123]

In partial fractions, why

\frac{3x+5}{(1-2x)^2} = \frac{A}{(1-2x)^2} + \frac{B}{(1-2x)}

and not
\frac{3x+5}{(1-2x)^2} = \frac{A}{(1-2x)} + \frac{B}{(1-2x)}

Why exists the exponent on the denominator in the right hand side of the equation?

 

[DonAntonio]

In partial fractions, why

\frac{3x+5}{(1-2x)^2} = \frac{A}{(1-2x)^2} + \frac{B}{(1-2x)}

and not
\frac{3x+5}{(1-2x)^2} = \frac{A}{(1-2x)} + \frac{B}{(1-2x)}

Why exists the exponent on the denominator in the right hand side of the equation?

Well, \frac{A}{(1-2x)} + \frac{B}{(1-2x)}=\frac{A+B}{1+2x}=\frac{C}{1+2x} , with C a consant, and this clearly cannot equal the original

expression since then we don't have a square in the denominator...

DonAntonio

 

[xeon123]

1 - So, why \frac{B}{1-2x} doesn't have an exponent of 2 in the denominator (\frac{B}{(1-2x)^2})?

2 - I also have a basic question related to your reply. Why
\frac{A}{(1-2x)} + \frac{B}{(1-2x)} = \frac{A+B}{(1+2x)} and not \frac{A+B}{(1-2x)}

 

[DonAntonio]

1 - So, why \frac{B}{1-2x} doesn't have an exponent of 2 in the denominator (\frac{B}{(1-2x)^2})?


*** Things get in order when you add this fraction to the first one, which has a square in the denominator, and then you

do common denominator to both fractions. ****



2 - I also have a basic question related to your reply. Why
\frac{A}{(1-2x)} + \frac{B}{(1-2x)} = \frac{A+B}{(1+2x)} and not \frac{A+B}{(1-2x)}

That was only a mistake of mine: I wrote + instead of - .

DonAntonio

 

[xeon123]

I'm trying to solve the equation, but I get different values:

\frac{A}{(1-2x)^2} + \frac{B}{(1-2x)} = \frac{A(1-2x)}{(1-2x)^3} + \frac{B(1-2x)^2}{(1-2x)^3} = \frac{(1-2x)(A+B(1-2x))}{(1-2x)^3} = \frac{A+B(1-2x)}{(1-2x)^2}

from this point forward I can't put this fraction to be like \frac{A+B}{(1-2x)}, I think...

 

[DonAntonio]

I'm trying to solve the equation, but I get different values:

\frac{A}{(1-2x)^2} + \frac{B}{(1-2x)} = \frac{A(1-2x)}{(1-2x)^3} + \frac{B(1-2x)^2}{(1-2x)^3} = \frac{(1-2x)(A+B(1-2x))}{(1-2x)^3} = \frac{A+B(1-2x)}{(1-2x)^2}

from this point forward I can't put this fraction to be like \frac{A+B}{(1-2x)}, I think...

Of course you can't. What makes you think you could? The original expression is a rational function

with denominator's degree equal to 2, and in the last line above you wrote a rational fraction with denominator's

degree equal to 1...of course they can't be equal.

DonAntonio

 

[xeon123]

So, how can both expressions be equal?

\frac{A}{(1-2x)^2} + \frac{B}{(1-2x)} = \frac{A+B}{(1-2x)}?

And, how do I know that
\frac{A}{(1-2x)^2} + \frac{B}{(1-2x)} = \frac{3x+5}{(1-2x)^2} ?

 

[DonAntonio]

So, how can both expressions be equal?

\frac{A}{(1-2x)^2} + \frac{B}{(1-2x)} = \frac{A+B}{(1-2x)}?}


*** Who said they are? ***


And, how do I know that
\frac{A}{(1-2x)^2} + \frac{B}{(1-2x)} = \frac{3x+5}{(1-2x)^2} ?

You can use the procedure to decompose in partial fractions: you have to know that it is possible to write

\displaystyle{\frac{3x+5}{(1-2x)^2} =\frac{A}{1-2x}+\frac{B}{(1-2x)^2}\,,\,\,A\,,\,B\,\,} constants. Now multiply through by the common denominator:

\displaystyle{ 3x+5=A(1-2x)+B\,\,} . This is a polynomial identity so you can compare particular values of x in both sides and/or

coefficients of powers of x to find out the values of the constants A,B. For example:

The coefficient in both sides of \,\,\displaystyle{x: 3=-2A\Longrightarrow\,A=-\frac{3}{2}} , and now the free coefficient

in both sides: \displaystyle{5=A+B\Longrightarrow B=5-A=5-\left(-\frac{3}{2}\right)=\frac{13}{2}\,\,} , so

\displaystyle{\frac{3x+5}{(1-2x)^2}=-\frac{3}{2(1-2x)}+\frac{13}{2(1-2x)^2}} , which you can easily check...

DonAntonio

 

[HallsofIvy]

Have you ever actually taken an algebra course? If so you should be able to do the sum on the left and see that it does, in fact, give the right side.

 

[Steely Dan]

In partial fractions, why

\frac{3x+5}{(1-2x)^2} = \frac{A}{(1-2x)^2} + \frac{B}{(1-2x)}

and not
\frac{3x+5}{(1-2x)^2} = \frac{A}{(1-2x)} + \frac{B}{(1-2x)}

Why exists the exponent on the denominator in the right hand side of the equation?

As DonAntonio pointed out, the reason it's not the second one is because the second one doesn't work. The next section on Wikipedia describes the general decomposition:

http://en.wikipedia.org/wiki/Partial_fraction#Over_the_reals

In particular, if you work it out you should be able to see that a repeated factor in the denominator results in the higher powers being seen in the decomposition.

 

[xeon123]

Thanks, now I got it.