To find $\displaystyle B$, after you have substituted the values for $\displaystyle A$ and $\displaystyle C$, you will have one equation in one unknown ($\displaystyle B$).
Expand everything and equate like powers of $\displaystyle x$.
What I have found is, since in this case B is made to be a product of X, we can substitute x=1 and sub values of what a and c is found to be. then we can find b without using the comparing method.
And in my opinion substituting doesn't fail! else my teacher shouldn't teach such methods which might not work sometimes...
You need to stop worrying about "rules", such as putting in specific numbers, and more about thinking about a problem. That was why Prove It kept asking you to give a specific example- you don't follow any fixed "algorithm", you do whatever is necessary for a specific problem.
If you have something like $\displaystyle \frac{x- 3}{(x-1)^3(x^2+ 4)}$, you can write that as $\displaystyle \frac{x- 3}{(x-1)^3(x^2+ 4)}= \frac{A}{x- 1}+ \frac{B}{(x-1)^2}+ \frac{C}{(x-1)^3}+ \frac{Dx+ E}{x^2+4}$.
You should start by multiplying both sides by the denominator so you get $\displaystyle x- 3= A(x-1)^2(x^2+4)+ B(x-1)(x^2+4)+ C(x^2+4)+ (Dx+ E)(x-1)^3$
Now there are two fundamentally different methods you could use to find A, B, C, D, and E. They both use the basic idea that to solve for 5 unknown values, we need 5 equations.
1) Multiply out the right sides and set "coeficients of like powers" equal.
$\displaystyle x- 3= Ax^4- 2Ax^3+ 5Ax^2- 8Ax+ A+ Bx^3- Bx^2+ 4Bx- 4B+$$\displaystyle Cx^2+ 4C+ Dx^4- 4Dx^3+ 3Dx^2- Dx+ Ex^3- 3Ex^2+ 3Ex- E$
$\displaystyle x- 3= (A+ D)x^4 +(-2A+ B+ E)x^3+ (5A- B+ C+ 3D- 3E)x^2+$$\displaystyle (-8A+4B- D+ 3E)x+ (A- 4B+ 4C- D- E)$
So we must have
A+ C= 0 since there is no "$\displaystyle x^4$" on the left
-2A+ B+ E= 0 since there is no "[tex]x^3[tex]" on the left
5A- B+ C+ 3D- 3E= 0 since there is no "$\displaystyle x^2$" on the left
-8A+ 4B+ D+ 3E= 1 since there is "x" on the left
A- 4B+ 4C- D- E= -3 since there is "-3" on the left
Solve those 5 equations for A, B, C, D, and E.
2) Get 5 equations by setting x equal to 5 different values in $\displaystyle x- 3= A(x-1)^2(x^2+4)+ B(x-1)(x^2+4)+ C(x^2+4)+ (Dx+ E)(x-1)^3$. IF a certain value sets a factor equal to 0 that will give a simpler equation but that is NOT necessary.
Since x= 1 makes (x-1)= 0, we certainly should try x= 1
$\displaystyle 1- 3= -2= A(0)+ B(0)+ C(1+ 4)+ (Dx+E)(0)= 5C$
we can solve that immediately for C= -2/5.
There is no other value of x that makes a factor 0 so just use whatever 4 numbers you want- preferably simple numbers like, say, 0, -1, -2, 2:
With x= 0,
$\displaystyle 0- 3= -3= 4A- 4B+ 4C- E$
With x= -1
$\displaystyle -1-3= -4= 20A-10B+ 5C- 8D+ 8E$
With x= 2
$\displaystyle 2-3= -1= 8A+ 8B+ 8C+ 2D+ E$
With x= -2
$\displaystyle -2-3= -5= 72A- 24B+ 8C- 16D- 8E$
Again, five equations to solve for A, B, C, D, and E.
Okay, for example, in this case(the question I posted earlier), provided me an x value to substitute x as, and that is (-3/4) but that would enable me to only find A. So I need to sub 0 and 1 to find unknowns C and B respectively.
So, my question is, what should I substitute as x after I have no more values to substitute x as? 0 and 1 in any case? up to 2, 3, 4 or 5 if i still cant find the unknown?
Substitute whatever values of x give you equations that you can use to solve for the unknowns.
You need to go back and review examples of partial fraction decomposition, this thread has given you as much help as it can and things are just starting to go around in circles.
Thread closed.