partial fractions main dont understand 1.?

Question

The following equation is the expression of the left hand side as a sum of partial fractions with rational numbers as coefficients. On the right hand side, A and B are constants and the numerator of the third term is missing

x^3−14x^2+26x−11/ = A/ . . . +B/ . . . . . . C/
. . (x−2)^2(x^2+3). . . .(x−2) . (x−2)^2 . (x^2)+3.

1.Supply the numerator of the last term giving appropriate names to the coefficients.
2.Determine the coefficient B.
3.Determine the remaining coefficients, including those in the missing numerator.

Answer 1

RHS = A/(x-2) + B/(x-2)^2 + (Cx+D)/(x^2 + 3)

This is the correct partial fraction formulation

Multiply numerator and denominator by (x-2)^2 (x^2 + 3)
x^3 - 14x^2 + 26x - 11
= A*(x-2)*(x^2 + 3) + B*(x^2 + 3) + (Cx+D)*(x-2)^2

Now to find those coefficients
Set x = 2
LHS = -7, RHS = 7B
B = -1

Equating coefficients:
x^3: A + C = 1
or C = 1-A

constant: -6A + 3B + 4D = -11
or -6A + 4D = -8
or 4D = 6A - 8

x: 3A + 4C - 4D = 26
or 3A + 4 - 4A + 8 - 6A = 26
A = -2, C = 3, D = -5

Final answer:
-2/(x-2) - 1/(x-2)^2 + (3x-5)/(x^2 + 3)

Answer 2

The proper decomposition is
= A/(x - 2) +B/(x - 2)^2 + (Cx + D)/(x^2 + 3) .

You should know how to find A = -2 , B = -1 , C = 3 , and D = -5.

Answer 3

when you do (A/2n-a million) + (B/2n+a million) you will possibly desire to equivalent it to (6) / ((2n-a million)(2n+a million)) so we've (2nA+B+2nB-B=6) (A+B)2n+(A-B)=6 so now with the aid of fact interior the the terrific option we've not got any n (A+B=0) and A-B=6 as quickly as we remedy those we whould have A=3 B=-3 in case you examine it you will locate the actuality