Please help I’m disliking algebra 2!!!

Answer:
[tex]C.\:h(x)=2(x-9)(x+3)^2[/tex]
Step-by-step explanation:
If the polynomial function has zeros at 9 with multiplicity 1, then [tex](x-9)[/tex] is a factor of the polynomial.
Also if the polynomial function has a zero at [tex]-3[/tex] with a multiplicity of 2, then
[tex](x+3)^2[/tex] is another factor of the polynomial.
The polynomial function can therefore be written as;
[tex]h(x)=a(x-9)(x+3)^2[/tex]
But the given polynomial has an [tex]a[/tex] value of 2, hence the required polynomial is
[tex]h(x)=2(x-9)(x+3)^2[/tex]