what is the value of A, B, and C of the partial fraction decomposition of this rational function.

(1x^2-1x+4)/((x+5)(x^2+9))= A/(x+5)+(Bx+C)/(x^2+9)

thanks

Printable View

- Feb 6th 2008, 10:35 AMwaite3partial fraction decomposition solve A,B and C
what is the value of A, B, and C of the partial fraction decomposition of this rational function.

(1x^2-1x+4)/((x+5)(x^2+9))= A/(x+5)+(Bx+C)/(x^2+9)

thanks - Feb 6th 2008, 10:29 PMCaptainBlack
You need to find $\displaystyle A,B$ and $\displaystyle C$ such that:

$\displaystyle A(x^2+9)+(Bx+C)(x+5) = x^2-x+4$

You do this by expanding the left hand side collecting the like powers of $\displaystyle x$ and then equating coefficients with those on the right hand side, so:

$\displaystyle A+B=1$

$\displaystyle 5B+C=-1$

$\displaystyle 9A+5C=4$

RonL