Question

Please help I’m disliking algebra 2!!!

Answer 1

Answer:

$C.\:h(x)=2(x-9)(x+3)^2$

Step-by-step explanation:

If the polynomial function has zeros at 9 with multiplicity 1, then $(x-9)$ is a factor of the polynomial.

Also if the polynomial function has a zero at $-3$ with a multiplicity of 2, then

$(x+3)^2$ is another factor of the polynomial.

The polynomial function can therefore be written as;

$h(x)=a(x-9)(x+3)^2$

But the given polynomial has an $a$ value of 2, hence the required polynomial is

$h(x)=2(x-9)(x+3)^2$