SOLUTION: partial fraction decomposition; use the partical fraction decomposition to "decompose" the following functions: 1) (x-8)/(x^2-3x-28) 2) (-x)/(x^2+3x+2)

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: partial fraction decomposition; use the partical fraction decomposition to "decompose" the following functions: 1) (x-8)/(x^2-3x-28) 2) (-x)/(x^2+3x+2)      Log On


   

Question 36698: partial fraction decomposition; use the partical fraction decomposition to "decompose" the following functions:
1) (x-8)/(x^2-3x-28)
2) (-x)/(x^2+3x+2)
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
partial fraction decomposition; use the partical fraction decomposition to "decompose" the following functions:
1) (x-8)/(x^2-3x-28)...FIRST FACTORISE THE D.R..
X^2-7X+4X-28=X(X-7)+4(X-7)=(X-7)(X+4)...HENCE LET
{(X-8)/(X-7)(X+4} = {A/(X-7)}+{B/(X+4)}={A(X+4)+B(X-7)}/(X-7)(X+4)..HENCE
X-8 = A(X+4)+B(X-7)....PUT X=-4 TO FIND B AND X=7 TO FIND A
X=-4.......-4-8=A*0+B(-4-7)=-11B....B=12/11
X=7.......7-8=A(7+4)+B*0=11A........A=-1/11
HENCE WE GET
(x-8)/(x^2-3x-28) = {-1/11(X-7)}+{12/11(X+4)}
2) (-x)/(x^2+3x+2)..PROCEEDING AS ABOVE
X^2+2X+X+2=X(X+2)+1(X+2)=(X+1)(X+2)
{-X/(X+1)(X+2)}= {A/(X+1)}+{B/(X+2)}
-X=A(X+2)+B(X+1)
PUT X=-1.......1=A(-1+2)+B*0=A.......A=1
PUT X=-2.......2=A*0+B(-2+1)=-B......B=-2
HENCE WE GET
(-x)/(x^2+3x+2)={1/(X+1)}-{2/(X+2)}