# Homework Help/Verify solution

• December 9th 2007, 04:16 PM
Doris McRae
Homework Help/Verify solution
Would very much appreciate verification of solution to partial fraction decomposition question.

Find the partial fraction decomposition of x^4/(x-1)^3

Since degree of numerator is greater than denominator ... long division gives
3x^3-3x^2+x. Since degrees are same, another long division gives:
6x^2 -8x +3.

With A/(x-1) + B/(x-1)^2 + C/(x-1)^3

My solution is: A=6, B=4 and C=1

But my "check" does not compute.

Appreciate any help or redirection.

Doris
• December 9th 2007, 04:28 PM
Krizalid
I'm not a close friend of partial fractions but I'll do this as follows:

$\frac{{x^4 }}
{{(x - 1)^3 }} = x \cdot \left( {\frac{x}
{{x - 1}}} \right)^3 .$

Now

$\frac{x}
{{x - 1}} = \frac{{x - 1 + 1}}
{{x - 1}} = 1 + \frac{1}
{{x - 1}}.$

And

$\left( {\frac{x}
{{x - 1}}} \right)^3 = 1 + \frac{3}
{{x - 1}} + \frac{3}
{{(x - 1)^2 }} + \frac{1}
{{(x - 1)^3 }}.$

Multiply this last result by $x$ and you'll get the desired partial fraction.

(Of course someone else will be give you a solution which involves your procedure.)
• December 10th 2007, 02:28 AM
Quote:

I'm not a close friend of partial fractions but I'll do this as follows:

http://www.mathhelpforum.com/math-he...d059451a-1.gif

Now

http://www.mathhelpforum.com/math-he...063ecbb4-1.gif

And

http://www.mathhelpforum.com/math-he...e859aaa8-1.gif

Multiply this last result by http://www.mathhelpforum.com/math-he...155c67a6-1.gif and you'll get the desired partial fraction.

(Of course someone else will be give you a solution which involves your procedure.)
THAT WAS PURE AWESOME

Hmmm... now I need to say something relevent to stop this post being spam.
Quote:

But my "check" does not compute.
I think that the problem is in your checking because my check worked fine:

A(x-1)^2+B(x-1)+C = 6x^2-8x+3
6(x^2-2x+1)+4(x-1)+1 = 6x^2-8x+3
6x^2-12x+6+4x-4+1 = 6x^2-8x+3
6x^2-8x+3 = 6x^2-8x+3