To find a degree 3 polynomial with zeros at $z = -1$, $z = -5$, and $z = -8$ and a leading coefficient equal to 1, we can use the fact that the polynomial can be written as:

$f(z) = a(z + 1)(z + 5)(z + 8)$

Given that the leading coefficient is 1, we set $a = 1$. Therefore, the polynomial is:

$f(z) = (z + 1)(z + 5)(z + 8)$

Now, we expand this expression to find the polynomial.

Step 1: Expand $(z + 5)(z + 8)$

$(z + 5)(z + 8) = z^2 + 8z + 5z + 40 = z^2 + 13z + 40$

Step 2: Expand $(z + 1)(z^2 + 13z + 40)$

$f(z) = z(z^2 + 13z + 40) + 1(z^2 + 13z + 40)$

$f(z) = z^3 + 13z^2 + 40z + z^2 + 13z + 40$

$f(z) = z^3 + 14z^2 + 53z + 40$

So, the polynomial is:

$f(z) = z^3 + 14z^2 + 53z + 40$

Would you like more details on this process or have any questions?

Here are some related questions for further exploration:

1. How would the polynomial change if the leading coefficient were different from 1?
2. What happens if one or more of the zeros is a repeated root?
3. How can you factor a degree 3 polynomial given its roots?
4. How do the coefficients of the polynomial relate to the sum and product of its roots?
5. How can you verify that the given polynomial has the correct roots?

Tip: Remember that the roots of a polynomial are closely related to its factors. Each root $r$ corresponds to a factor of the form $(z - r)$.