Respuesta :

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]

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