SOLUTION
From the roots of the eqution
[tex]\begin{gathered} x=-3 \\ x=2 \\ x=5 \end{gathered}[/tex]We have
[tex](x+3)(x-2)(x-5)[/tex]So the form should be
[tex]\begin{gathered} y=a(x+3)(x-2)(x-5)_{} \\ \\ \text{Where a is a constant } \end{gathered}[/tex]Now, let's find a. From
[tex]\begin{gathered} y=a(x+3)(x-2)(x-5) \\ \\ 3\times-2\times-5=30 \end{gathered}[/tex]Since y = -6, we have that
[tex]\begin{gathered} -6=30\times a \\ -6=30a \\ a=\frac{-6}{30} \\ \\ a=-\frac{1}{5} \end{gathered}[/tex]The formula for p(x) becomes
[tex]y=-\frac{1}{5}(x+3)(x-2)(x-5)[/tex]