Since the zeros of the function are (3, 0) and (-5, 0), we can use the value of p = 3 and q = -5 in the factored form.
Also, in order to find the value of a, we can use the ordered pair (-1, 16) in the function, so we have:
[tex]\begin{gathered} y=a(x-3)(x+5)_{} \\ (-1,16)\colon \\ 16=a(-1-3)(-1+5) \\ 16=a(-4)(4) \\ 16=-16a \\ a=\frac{16}{-16} \\ a=-1 \end{gathered}[/tex]
So the factored form of the equation is:
[tex]y=-1(x-3)(x+5)[/tex]
