Expressing a rational expression in partial fractions

[MathsRetard09]

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, I am 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 I am looking for a bit of detail involved, if you cba its ok, like i say I am helping someone else, but trying to understand at the same time.

Thanks :)

 

[tiny-tim]

Hi MathsRetard09!

(try using the X² icon just above the Reply box )

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) ​

 

[HallsofIvy]

I assume you mean
$$\frac{x^2- x+ 1}{(x-2)(x^2+ 1)}= \frac{A}{x- 2}+ \frac{Bx+ C}{x^2+ 1}$$.

Multiply on both sides by $(x- 2)(x^2+ 1)$ to get
$$x^2- x+ 1= A(x^2+ 1)+ (Bx+ C)(x- 2)$$

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
$$x^2- x+ 1= A(x^2+ 1)+ (Bx+ C) (x-2)= Ax^2+ A+ Bx^2+ (C- 2B)x- 2C$$$$= (A+ B)x^2+ (C- 2B)x+ (A- 2C)$$

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 $x^2+ 1$, 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.