Let's see if you "know the methods". How did you set it up?
Again with the bad notation. This may be part of the problem you are having.
I'm sure you mean: (x-2) = (ax+b)(x-1)^2 +c (x^2+1)(x-1) +d(x^2+1)
1) I do not understand the use of x = -1. Don't you mean +1?When x = -1
2) x = 1 is NOT in the Domain. How can you substitute it? I realize this is a popular method for this sort of thing. I never have liked it for the reason just stated.
The solution is in Algebra. No need for a trick that works most of the time but really makes no sense.
Expand and collect on the right-hand side to get .
Matching corresponding powers of x in the numerators gives
a+c = 0
-2a+b-c+d = 0
a-2b+c = 1
b-c+d = -2
You must solve this system.
If you already have one of the values from somewhere else, you may as well use it.
2) x = 1 is NOT in the Domain. How can you substitute it? I realize this is a popular method for this sort of thing. I never have liked it for the reason just stated.
currently i am studying in lower6, as i said , that is the frm the example tht is written from the book.
Well, what can I say? Some examples and methods are more harm than they are help.
Others may disagree with me. Obviously your author does. Soroban does. For me, just don't use it. Run away screaming. Trust me on this. Use algebra rather than trickery, every time.
My views. I welcome others'.
What you are saying is that your choices are:
1) Use the trick, knowing that it doesn't work all the time and that it has a shaky theoretical basis, or
2) Fail.
That is not a good idea.
The solution is not difficult. Just be careful and deliberate.
After substituting d = -1/2 and a = -c you are left with:
b + c = 1/2 and b = -1/2
This makes c = 1 and a = -1.
Where is the confusing part? Careful and deliberate. You aren't trusting me, yet. {encourage}{encourage}{encourage}