Respuesta :

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]

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