\(\dfrac{x^3+ax+b}{x-2}=\dfrac{x^3-2x^2+2x^2-4x+\left(a+4\right)x-2a-8+2a+8+b}{x-2}\)
\(=x^2+2x+\left(a+4\right)+\dfrac{2a+b+8}{x-2}\)
=>2a+b+8=0
\(\dfrac{x^3+ax+b}{x+1}=\dfrac{x^3+x^2-x^2-x+\left(a+1\right)x+a+1+b-a-1}{x+1}\)
\(=x^2-x+\left(a+1\right)+\dfrac{b-a-1}{x+1}\)
=>-a+b-1=0
mà 2a+b+8=0
nên a=-3; b=-2
=>3a-2b=3*(-3)-2*(-2)=-9+4=-5