Question 3.3.2: Factoring a Polynomial Given a Zero Factor ƒ(x) = 6x³+ 19x² ...

Because -3 is a zero of ƒ, x – (-3) = x + 3 is a factor.

\begin{matrix} -3) \overline{\begin{matrix} 6 & 19 & 2&-3\end{matrix} } &\text{Use synthetic division to}\\ \underline{\begin{matrix} & &-18& -3&3\end{matrix} } &\text{divide ƒ(x) by}  x + 3.\\ \begin{matrix}& 6 & 1 &-1&0\end{matrix} \end{matrix}

The quotient is 6x² + x – 1, which is the factor that accompanies x + 3.

\begin{matrix} ƒ(x) = (x + 3)(6x² + x – 1)\\ ƒ(x)=\underbrace{(x + 3)(2x + 1)(3x – 1)}_{\text{These factors are all linear.}} &\text{Factor}   6x² + x – 1. \end{matrix}