From the table, we were given the zero of the function h(x).
The zeros of the function are
[tex]-3,-2,\text{ and 1}[/tex]
The formula to obtain the equation of the function is,
[tex](x-a)(x-b)(x-c)[/tex]
Where
[tex]\begin{gathered} a=-3 \\ b=-2 \\ c=1 \end{gathered}[/tex]
Hence,
[tex]\begin{gathered} h(x)=(x--3)(x--2)(x-1) \\ h(x)=(x+3)(x+2)(x-1) \end{gathered}[/tex]
Expanding the function above
[tex]\begin{gathered} h(x)=x(x+2)+3(x+2)(x-1) \\ h(x)=x^2+2x+3x+6(x-1) \\ h(x)=x^2+5x+6(x-1) \\ h(x)=x(x^2+5x+6)-1(x^2+5x+6) \\ h(x)=x^3+5x^2+6x-x^2-5x-6 \\ h(x)=x^3+5x^2-x^2+6x-5x-6 \\ h(x)=x^3+4x^2+x-6 \end{gathered}[/tex]
Therefore, the equation of the polynomial function is
[tex]h(x)=x^3+4x^2+x-6[/tex]
Hence, the answer is Option 3.
