how can i do partial fraction for this equation
2/(x^2 +.786x + .212)
plz help
what will happend if there is another term like this in denominator
2/((x^2 +.786x + .212)(x^2+.23x+.9))
thanks
For the first one since the discriminant is negative
$\displaystyle b^2-4ac=(.786)-4(1)(.212) < 0$ the quadratic is irreducible.
So over the real numbers there is not a partial fraction decomposition.
Again fro the 2nd one both quadratic factors are irreducible we get that
$\displaystyle \dipslaystyle \frac{Ax+B}{x^2 +.786x + .212}+\frac{Cx+D}{x^2+.23x+.9}=\frac{2}{(x^2 +.786x + .212)(x^2+.23x+.9)}$
Now multiply the LCD to get
$\displaystyle \displaystyle (Ax+B)(x^2+.23x+.9)+(Cx+D)(x^2 +.786x + .212)=2$
Now expand everything out and collect all of the x's to get
$\displaystyle (A+C)x^3+(.23A+B+.786C+D)x^2+(.9A+.23B+.212C+.786D )x+(.9B+.212D)=2=0x^3+0x^2+0x+2$
Now equate the coefficients to get a system of 4 equations in 4 unknowns.