a: f(x)=x3+x-2
\(\dfrac{f\left(x\right)}{g\left(x\right)}=\dfrac{x^3+x-2}{x^2-x-2}\)
\(=\dfrac{x^3-x^2-2x+x^2-x-2+4x}{x^2-x-2}=x+1+\dfrac{4x}{x^2-x-2}\)
c: Để f(x) chia hết cho g(x) thì \(x^3+ax+b⋮x^2-x-2\)
\(\Leftrightarrow x^3-x^2-2x+x^2-x-2+\left(a+3\right)x+b+2⋮x-2\)
=>a=-3 và b=-2