Question 1096768
<pre>
Given zeros:

4, -14, and 5 + 8i

Let p(x) be the polynomial. 
Since 5+8i is a zero of p(x), and the polynomial has real coefficients
then so is its conjugate 5-8i, so four of the zeros of p(x) are

4, -14, 5+8i and 5-8i, so the solutions to the equation p(x)=0 are: 

{{{x}}}{{{""=""}}}{{{4}}}, {{{x}}}{{{""=""}}}{{{-14}}}, {{{x}}}{{{""=""}}}{{{5+8i}}}, {{{x}}}{{{""=""}}}{{{5-8i}}}

Get 0 on the right of those four equations:

{{{x-4}}}{{{""=""}}}{{{0}}}, {{{x+14}}}{{{""=""}}}{{{0}}}, {{{x-5-8i}}}{{{""=""}}}{{{0}}}, {{{x-5+8i}}}{{{""=""}}}{{{0}}}

Multiply the left and right sides

{{{(x-4)(x+14)(x-5-8i)(x-5+8i)}}}{{{""=""}}}{{{(0)(0)(0)(0)}}}

{{{(x-4)(x+14)(x-5-8i)(x-5+8i)}}}{{{""=""}}}{{{0}}}

If we want p(x) to have leading coefficient 1, 
and have only those four zeros and no others, then
it will have leading coefficient x<sup>4</sup> and
be of degree 4.

{{{p(x)}}}{{{""=""}}}{{{(x-4)(x+14)(x-5-8i)(x-5+8i)}}}
 
Now you'll have to multiply that out all by yourself,
which is a big job.  But when you finish you'll get:

{{{p(x)}}}{{{""=""}}}{{{x^4 - 67x^2 + 1450x - 4984}}}

Edwin</pre>