SOLUTION: I haven't done this before. I'm so confused. N=3 -2 and 8+5i are zeros. F(2)=244. Says find the nth degree polynomial function with real coefficients satisfying the given condition

Apply these two rules:

1. Each zero b contributes a factor  (x-b)

2. Complex zeros (and roots) always come in conjugate pairs





The zero at 3  contributes  (x-3)

The zero at -2 contributes  (x-(-2)) = (x+2)

The zero at 8+5i and conjugate contribute  (x-8-5i)(x-8+5i)



Multiply all four of these factors, and simplify, to get:



 





But, this function has  g(2) = -244, and the problem states f(2)=244,

thus we need to multiply g(x) by -1, and the final answer is:



  



Note that f(x) is just g(x) flipped across the x-axis (mirror image).

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EDIT 12/1 --  I see (now, after seeing MathTherapy's answer) that the intention was to indicate N=3 as the _DEGREE_ of the polynomial.  I interpreted N=3 as one of the zeros.   Thanks MathTherapy for catching that.  



 

N = 3 indicates that there are 3 zeroes, or 3 solutions. One of the 3 zeroes or solutions is - 2, and the other 2 are:
8 + 5i, and its CONJUGATE, 8 - 5i. 

With the 3 zeroes/solutions being - 2, 8 + 5i, and 8 - 5i, it follows that, x = - 2, x = 8 + 5i, and x = 8 - 5i, thus 
making the function’s factors: x + 2, x - 8 - 5i, and x - 8 + 5i.

We now have the following function: f(x) = a(x + 2)(x - 8 - 5i)(x - 8 + 5i), which then becomes:
     y = a(x + 2)[(x - 8)2 - (5i)2] ------ FOILing/Expanding (x - 8 - 5i)(x - 8 + 5i)
     
     
   -------- Substituting f(2) = 244, or (2, 244) for (x, y) to determine value of “a”


With ”a” being 1, 

This is your 3rd degree polynomial, or a polynomial with N = 3, and the given roots.