Since the parabola crosses the $x$-axis at $x=2$ and $x=8$, its equation can be written in factored form as $y=a(x-2)(x-8)$ for some constant $a$. Using the point where it crosses the $y$-axis, $(0,32)$, substitute $x=0$ and $y=32$:

So $a=2$, and the equation is

The area function is

This is a quadratic that opens downward, so its maximum value occurs at the vertex. For

the vertex has $x$-coordinate

Then the maximum area is

So the maximum possible area is $200$ square feet.

Because the remainders on division by $x-2$ and $x+1$ are known, the Remainder Theorem gives $p(2)=5$ and $p(-1)=-4$. If $(x-2)(x+1)$ is factored out, the remaining part must be a linear expression, since any polynomial can be written in the form

Then substituting the two given values gives

and

Solving this system: subtract the second equation from the first to get $3a=9$, so $a=3$. Then $-3+b=-4$, so $b=-1$. Therefore

which matches the correct expression.

Because the graph has intercepts at $t=0$ and $t=6$, the function must include factors $t$ and $(t-6)$. So $h(t)$ can be written as $at(t-6)$. Using the point $(2,32)$ gives

so $a=-4$. Therefore,

This function also makes sense for the situation because the parabola opens downward, so the ball rises and then falls back to the ground.

Use the Remainder Theorem and the given factors. Since the remainder when dividing by $x-1$ is $6$, $p(1)=6$. Since division by $x+2$ leaves remainder $0$, $p(-2)=0$, so $x+2$ is a factor. Also, $x-3$ is a factor, so $p(3)=0$. Because the polynomial is monic and cubic, it must be

for some number $r$. Now use $p(1)=6$:

So

Thus

To find $a+b+c$, expand:

so

Therefore, $a=-3$, $b=-4$, and $c=12$, and