\(\text{Ta có : }f_{\left(x\right)}=g_{\left(x\right)}\cdot Q_{\left(x\right)}+R\\ \Rightarrow3x^3-2x^2+5=g_{\left(x\right)}\left(3x-2\right)+\left(3x+3\right)\\ \Rightarrow g_{\left(x\right)}\left(3x-2\right)=\left(3x^3-2x^2+5\right)-\left(3x+3\right)\\ \Rightarrow g_{\left(x\right)}\left(3x-2\right)=3x^3-2x^2-3x+2\\ \Rightarrow g_{\left(x\right)}=\left(3x^3-2x^2-3x+2\right):\left(3x-2\right)\\ \Rightarrow g_{\left(x\right)}=\left[\left(3x^3-2x^2\right)-\left(3x-2\right)\right]:\left(3x-2\right)\\ \Rightarrow g_{\left(x\right)}=\left[x^2\left(3x-2\right)-\left(3x-2\right)\right]:\left(3x-2\right)\\ \Rightarrow g_{\left(x\right)}=\left(x^2-1\right)\left(3x-2\right):\left(3x-2\right)\\ \Rightarrow g_{\left(x\right)}=x^2-1\\ Vậy\text{ }g_{\left(x\right)}=x^2-1 \)
