Expressing a rational expression in partial fractions

  • #1

Main Question or Discussion Point

Trying to help out a friend.

I appolagise if yet again this is in the wrong part of the forum, i haven't an idea what it is categorised as, im an apprentice engineer and simply that lol

Can someone explain how the following is done:

x^2 – x + 1 / (x - 2)(x^2 + 1) ≅ A / (x - 2) + Bx + C / x^2 + 1

"Using a mixture of substitution and equating gives A= 3/5, B= 2/5, C = -1/5 "

I truly would appreciate any replies, oddly enough i actually want to understand what this type of mathematics is used for so yes im looking for a bit of detail involved, if you cba its ok, like i say im helping someone else, but trying to understand at the same time.

Thanks :)
 

Answers and Replies

  • #2
tiny-tim
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Hi MathsRetard09! :smile:

(try using the X2 icon just above the Reply box :wink:)
x^2 – x + 1 / (x - 2)(x^2 + 1) ≅ A / (x - 2) + Bx + C / x^2 + 1

"Using a mixture of substitution and equating gives A= 3/5, B= 2/5, C = -1/5 "
Is there a bracket shortage where you are??

I assume you mean (x2 – x + 1) / (x - 2)(x2 + 1) ≅ A / (x - 2) + (Bx + C) / x2 + 1 …

just expand the RHS over the whole (x - 2)(x2 + 1) :smile:
 
  • #3
HallsofIvy
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I assume you mean
[tex]\frac{x^2- x+ 1}{(x-2)(x^2+ 1)}= \frac{A}{x- 2}+ \frac{Bx+ C}{x^2+ 1}[/tex].

Multiply on both sides by [itex](x- 2)(x^2+ 1)[/itex] to get
[tex]x^2- x+ 1= A(x^2+ 1)+ (Bx+ C)(x- 2)[/tex]

Now there are several ways to find A, B, and C.

If you multiply the terms on the right and "combine like terms" you get
[tex]x^2- x+ 1= A(x^2+ 1)+ (Bx+ C) (x-2)= Ax^2+ A+ Bx^2+ (C- 2B)x- 2C[/tex][tex]= (A+ B)x^2+ (C- 2B)x+ (A- 2C)[/tex]

Now, in order for two polynomials to be equal for all x, their corresponding coefficients must be equal:
A+ B= 1, C- 2B= -1, A- 3C= 1, three equations to solve for A, B, and C.

Or, just put any three numbers you like for x to get three equations. Because of that "x- 2" term, x= 2 makes an especially simple equation:
2^2- 2+ 1= 3= A(2^2+ 1)+ (2B+ C)(2- 2)= 5A so A= 3/5.

Since there is no x that will make [itex]x^2+ 1[/itex], just put, say, x= 0 and x= 1.
If x= 0, 0^2- 0+ 1= 1= A(0^2+ 1)+ (2B+ C)(0-2)= A- 4B- 2C and,
if x= 1, 1^2- 1+ 1= 1= A(1^2+ 1)+ (2B+ C)(1- 2)= 2A- 2B- 2C.

Since we already know A, that is two equations to solve for B and C.
 
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