Hallo,

es gilt:

\( f(x) = \frac {4x+12} {x^3 + 3x^2 -x -3} = \frac a {x-1} + \frac b {x+1} + \frac c {x+3} \\ \Rightarrow  \frac a {x-1} + \frac b {x+1} + \frac c {x+3} = \frac {a(x+1)} {(x-1)(x+1)} + \frac {b(x-1)}{(x-1)(x+1)} + \frac c {x+3} \\ = \frac {a(x+1) + b (x-1)} {x^2-1} + \frac {c(x^2-1} {(x+3)(x^2-1)} \\ = \frac {(ax+a+bx-b)(x+3)} {(x^2-1)(x+3)} + \frac {cx^2-c} {x^3+3x^2-x-3} = \frac {((a+b)x + (a-b))(x+3) + (cx^2-c)} {x^3 +3x^2 -x-3} = \frac {((a+b)x^2 + (4a+2b)x + (3a-3b)+(cx^2 - c)} {x^3 +3x^2 -x- 3} \\ = \frac {(a+b+c)x^2 + (4a-2b)x + (3a-3b-c)} {x^3 +3x^2 -x-3} \)

Nun muss gelten:

\(  \frac {4x+12} {x^3 + 3x^2 -x -3}= \frac {(a+b+c)x^2 + (4a-2b)x + (3a-3b-c)} {x^3 +3x^2 -x-3}  \)

Jetzt kannst du durch Koeffizientenvergleich \( a,b \) und \( c \) bestimmen.

Grüße Christian