Ta có

\(\begin{array}{l} \begin{array}{*{20}{l}} {{\mkern 1mu} {x^3} + 4{x^2} + x - 6 = 0{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} }\\ { \Leftrightarrow {x^3} - {x^2} + 5{x^2} - 5x + 6x - 6 = 0} \end{array}\\ \begin{array}{*{20}{l}} { \Leftrightarrow {x^2}\left( {x - 1} \right) + 5x\left( {x - 1} \right) + 6\left( {x - 1} \right) = 0}\\ { \Leftrightarrow \left( {x - 1} \right)\left( {{x^2} + 5x + 6} \right) = 0}\\ { \Leftrightarrow \left( {x - 1} \right)\left( {{x^2} + 2x + 3x + 6} \right) = 0{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} }\\ {{\mkern 1mu} {\mkern 1mu} \Leftrightarrow \left( {x - 1} \right)\left[ {x\left( {x + 2} \right) + 3\left( {x + 2} \right)} \right] = 0}\\ { \Leftrightarrow \left( {x - 1} \right)\left( {x + 2} \right)\left( {x + 3} \right) = 0.} \end{array}\\ \to \left[ \begin{array}{l} x = 1\\ x = - 2\\ x = - 3 \end{array} \right. \end{array}\)

Vậy \(S={1;−2;−3}\) nên tích các nghiệm là \(1.(−2).(−3)=6\)