Given an equation to solve:
[tex]48q^3+32q^2-27q-18=0[/tex]let us solve,
[tex]\begin{gathered} 48q^3+32q^2-27q-18=0 \\ \mleft(3q+2\mright)\mleft(4q+3\mright)\mleft(4q-3\mright)=0 \\ 3q+2=0 \\ 4q+3=0 \\ 4q-3=0 \end{gathered}[/tex][tex]\begin{gathered} 3q+2=0 \\ q=-\frac{3}{2} \end{gathered}[/tex][tex]\begin{gathered} \: 4q+3=0 \\ q=-\frac{3}{4} \end{gathered}[/tex][tex]\begin{gathered} 4q-3=0 \\ q=\frac{3}{4} \end{gathered}[/tex]The solutions are,
[tex]q=-\frac{2}{3},\: q=-\frac{3}{4},\: q=\frac{3}{4}[/tex]