On considère la fonction $f$ définie par :
\[
f(x)=3x^3+2x^2-3x-2.
\]
\begin{enumerate}
\item Montrer que $f(1)=0$.
\item En déduire une factorisation de $f(x)$ sous la forme $f(x)=(x-1)(ax^2+bx+c)$.
\item En déduire la courbe représentative de $f$ parmi les trois proposées ci-dessous.
\begin{center}
\begin{tikzpicture}[>=latex,yscale=0.7]
\node[fill=purple!20,text=purple] at (0,-5) {$a$};
\draw[thick,->] (-2,0) -- (2,0);
\draw[thick,->] (0,-4) -- (0,5);
\draw (-1,0.1) -- (-1,-0.1) node[below] {$-1$};
\draw (1,0.1) -- (1,-0.1) node[below] {$1$};
\foreach\y in {-4,-3,-2,-1,1,2,3}
{
	\draw (0.1,\y) -- (-0.1,\y) node[left] {$\y$};
}
\node[below left] at (0,0) {$0$};
\clip (-2,-4) rectangle (2,5);
\draw[thick,purple] plot[domain=-5:4,samples=100] (\x,{3*\x*\x*\x+2*\x*\x-3*\x-2});
\end{tikzpicture}
\qquad
\begin{tikzpicture}[>=latex,yscale=0.7]
\node[fill=purple!20,text=purple] at (0,-5) {$b$};
\draw[thick,->] (-2,0) -- (2,0);
\draw[thick,->] (0,-4) -- (0,5);
\draw (-1,0.1) -- (-1,-0.1) node[below] {$-1$};
\draw (1,0.1) -- (1,-0.1) node[below] {$1$};
\foreach\y in {-4,-3,-2,-1,1,2,3}
{
	\draw (0.1,\y) -- (-0.1,\y) node[left] {$\y$};
}
\node[below left] at (0,0) {$0$};
\clip (-2,-4) rectangle (2,5);
\draw[thick,purple] plot[domain=-5:4,samples=100] (\x,{-3*\x*\x*\x+2*\x*\x+3*\x+2});
\end{tikzpicture}
\qquad
\begin{tikzpicture}[>=latex,yscale=0.7]
\node[fill=purple!20,text=purple] at (0,-5) {$c$};
\draw[thick,->] (-2,0) -- (2,0);
\draw[thick,->] (0,-4) -- (0,5);
\draw (-1,0.1) -- (-1,-0.1) node[below] {$-1$};
\draw (1,0.1) -- (1,-0.1) node[below] {$1$};
\foreach\y in {-4,-3,-2,-1,1,2,3}
{
	\draw (0.1,\y) -- (-0.1,\y) node[left] {$\y$};
}
\node[below left] at (0,0) {$0$};
\clip (-2,-4) rectangle (2,5);
\draw[thick,purple] plot[domain=-5:4,samples=100] (\x,{(\x-1)*(\x*\x+\x+2)});
\end{tikzpicture}
\end{center}
\end{enumerate}

