tolong dibantu lagi:(

[tex]\displaystyle \frac{x^3+ax^2-4x-12}{x^3-bx^2-4x+8}=\frac{x^2(x+a)-4(x+3)}{x^2(x-b)-4(x-2)}\\\\\text{agar dapat dijadikan pecahan linear, maka }\\x+a\equiv x+3\wedge x-b\equiv x-2\\a\equiv3\wedge-b\equiv-2\\\boxed{\boxed{a\equiv3\wedge b\equiv2}}[/tex]