Question 1188217
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 conditions. 
<pre>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)<sup>2</sup> - (5i)<sup>2</sup>] ------ FOILing/Expanding (x - 8 - 5i)(x - 8 + 5i)
    {{{matrix(1,3, y, "=", a(x + 2)((x - 8)^2  -  25i^2))}}} 
    {{{matrix(1,3, y, "=", a(x + 2)((x - 8)^2  -  25(- 1)))}}} 
  {{{matrix(1,3, 244, "=", a(2 + 2)((2 - 8)^2 + 25))}}} -------- Substituting f(2) = 244, or (2, 244) for (x, y) to determine value of “a”
{{{matrix(5,3, 244, "=", a(4)((- 6)^2 + 25), 244, "=", a(4)(36 + 25), 244, "=", a(4)(61), 244/4(61), "=", a, 1, "=", a)}}}

With ”a” being 1, {{{highlight_green(system(matrix(1,3, f(x), "=", a(x + 2)((x - 8)^2 + 25)), "becomes:", highlight(matrix(1,3, f(x), "=", (x + 2)((x - 8)^2 + 25))), or, highlight(matrix(1,4, f(x), "=", x^3 - 14x^2 + 57x + 178, "(expanded)")))))}}}

This is your 3<sup>rd</sup> degree polynomial, or a polynomial with N = 3, and the given roots.</pre>