Question 1099067
<pre>

Since 1 + i is a solution, so is its conjugate 1 - i

Set x = to each and get 0 on the right of each equation:

  x = -4;       x = 1+i;         x = 1-i
x+4 = 0;    x-1-i = 0;       x-1+i = 0

Multiply left sides and right sides of the three equations:

      (x+4)(x-1-i)(x-1+i) = (0)(0)(0)

Multiply and simplify:

  (x+4)[(x-1)-i][(x-1)+i] = 0  

         (x+4)[(x-1)²-i²] = 0

Square the binomial and replace i² by -1

      (x+4)[x²-2x+1-(-1)] = 0

         (x+4)[x²-2x+1+1] = 0

           (x+4)[x²-2x+2] = 0 

       x³-2x²+2x+4x²-8x+8 = 0

              x³+2x²-6x+8 = 0   

So the polynomial Q(x) is

                     Q(x) = x³+2x²-6x+8    

Edwin</pre>