b: f(x)=3x^3+4x^2-2x+7
\(\dfrac{f\left(x\right)}{g\left(x\right)}=\dfrac{3x^3+4x^2-2x+7}{x+2}\)
\(=\dfrac{3x^3+6x^2-2x^2-4x+2x+4+3}{x+2}\)
=3x^2-2x+2+3/x+2
Số dư là 3
c: \(\dfrac{f\left(x\right)}{g\left(x\right)}=\dfrac{x^3\left(x-5\right)+2\left(x-5\right)}{x-5}=x^3+2\)
=>Số dư là 0