Since the graph crosses the $x$-axis at $x=-3$ and $x=5$, the quadratic must have the form

Using the point $(1,-16)$, substitute $x=1$ and $f(1)=-16$:

So $a=1$. Therefore,

The $y$-intercept is the value of the function when $t=0$. Substitute $t=0$ into $h(t)=-16t^2+48t+20$:

So the ball's height at the start is $20$ feet.

First factor the numerator:

Then the expression becomes

For all $x\ne 2$, the common factor $x-2$ can be divided out, leaving

So the equivalent expression is $x-3$.

Since the graph crosses the $x$-axis at $x=1$ and $x=5$, the polynomial can be written as

Using the point $(2,-6)$, substitute $x=2$ and $p(2)=-6$:

So $a=2$. Then

Therefore, the value of $p(0)$ is $10$.

By the Remainder Theorem, $p(2)=5$ and $p(-1)=-1$. Since $q(x)=p(x)-2x$, evaluate $q$ at those same inputs:

and

So $q(x)$ leaves a remainder of $1$ when divided by $x-2$ and also a remainder of $1$ when divided by $x+1$. That statement must be true.