Candidates for global maxima and global minima exist where the derivative equals $0$ AND where the derivative does not exist.... Notice that this is a function that has a closed domain. That means that we already know two candidates for global max and min: the endpoints.

Evaluate the endpoints:
$p(−1) = 9$
$p(5) = 21^² = 441$
These might be the global min and max, respectively. We still need to check the other candidates.

Find additional candidates by setting the derivative equal to zero, and solving for $x$.
$p'(x) = 4x(x^² − 4)$
$0 = 4x(x − 2)(x + 2)$
$x = 0$, $x = 2$ and $x = −2$ are candidates...
but eliminate $x = −2$ because it is not within the given domain.

Evaluate the remaining candidates: $p(0) = 16$, $p(2) = 0$ and compare to the endpoint candidates.

Recall that maxima and minima are $y$-values.
Therefore the global maximum is $441$ and the global minimum is $0$.It should also be noted that $16$ is a local maximum.
Confirm this by sketching $p(x)$ on your graphing calculator.