Question 1057047
Find a polynomial equation with real coefficients that has 
the roots of -5, 2+i
<pre>
Since 2+i is a solution, and since the coefficients are all real,
its conjugate 2-i is also a solution.  Begin with

     x = 5;  x = 2+i;  x = 2-i

Get 0 alone on the right side of each:

   x+5 = 0;  x-2-i = 0; x-2+i = 0

Multiply all left sides together [and right sides too! (0)(0)(0)=0]

      (x+5)(x-2-i)(x-2+i) = 0

Simplify:

  (x+5)[(x-2)-i][(x-2)+i] = 0
         (x+5)[(x-2)²-i²] = 0
       (x+5)[(x-2)²-(-1)] = 0
          (x+5)[(x-2)²+1] = 0
      (x+5)[(x-2)(x-2)+1] = 0
         (x+5)[x²-4x+4+1] = 0
           (x+5)(x²-4x+5) = 0
             x³+x²-15x+25 = 0
  
So those zeros would be obtained when we set the
function f(x) = x³+x²-15x+25 equal to 0.  So
one answer is: 

f(x) = x³+x²-15x-25

Other answers would be gotten by multiplying the
right side by different constants <font color="red"><b>k</b></font>:
f(x) = <font color="red"><b>k</b></font>x³+<font color="red"><b>k</b></font>x²+25<font color="red"><b>k</b></font>x-15<font color="red"><b>k</b></font>

Edwin</pre>