`ĐK:x\ne -3`
`A(x)=3x^{3}+5x^{2}+32`
`=(3x^{3}+9x^{2})-(4x^{2}+12x)+(12x+36)-4`
`=3x^{2}(x+3)-4x(x+3)+12(x+3)-4`
`=(x+3)(3x^{2}-4x+12)-4`
Để `A(x)` chia hết cho `B(x)`
Thì : \(4⋮\left(x+3\right)=>x+3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
`=>x\in {-4;-2;-5;-1;-7;1}\ (TMDK)`