Question

Third-degree, with zeros of 4-I,4+I and a leading coefficient of -5

Answer 1

Please use "i" for the imaginary operator, not "I."

A third degree poly would have 3 factors and 3 zeros; you've supplied only 2 zeros.  

Let's model the polynomial as y = ax^3 + bx^2 + cx + d.
We are told that the leading coefficient is -5.  Therefore, our model becomes

y = -5x^3 + bx^2 + cx + d

The two given roots are 4-i and 4+i; if either is subst. into the above equation, y takes on the value 0 (this is the meaning of "root")

Thus, if x = 4-i, 0 = -5(4-i)^3 + b(4-i)^2 + c(4-i) + d.
          If x = 4+i, 0 = -5(4+i)^3 + b(4+i)^2 + c(4+i) + d

Here we have two equations in three unknowns (b, c and d).  There's not enough info to enable us to determine all three b, c and d.

One thing you might do at this point would be to set d=0 arbitrarily, and then solve the resulting two equations (without d) for b and c.

Can you take the ball and run with it from this point on?