SOLUTION: Find a polynomial function of least degree having only real coefficients, a leading coefficient of 1, and zeros of 4, 1-3i, and 2+2i.

zeros: 4, 1-3i, and 2+2i

Since 1-3i is a zero, so is 1+3i
Since 2+2i is a zero, so is 2-2i

So we have

x=4;    x=1-3i;    x=1+3i;    x=2+2i;    x=2-2i

Get 0 on the right of each of those:

x-4=0;  x-1+3i=0;  x-1-3i=0;  x-2-2i=0   x-2+2i=0

x-4=0;  (x-1)+3i=0;  (x-1)-3i=0;  (x-2)-2i=0   (x-2)+2i=0

Put parentheses around the first two terms in the last 4

Multiply equals by equals to get the polynomial function p(x)

p(x) = (x-4)[(x-1)+3i][(x-1)-3i][(x-2)-2i][(x-2)+2i] = 0

p(x) = (x-4){[(x-1)+3i][(x-1)-3i]}{[(x-2)-2i][(x-2)+2i]} = 0

p(x) = (x-4){(x-1)2-9i2}{(x-2)2-4i2} = 0

p(x) = (x-4){(x-1)2-9(-1)}{(x-2)2-4(-1)} = 0

p(x) = (x-4){(x-1)2+9}{(x-2)2+4} = 0

p(x) = (x-4){(x-1)2(x-2)2+ 4(x-1)2+9(x-2)2+36} = 0

p(x) = (x-4){(x-1)2(x-2)2+ 13(x-1)2+36} = 0
...

Keep on multiplying out and collecting like terms and 
you'll end up with this:

p(x) = x5-10x4+50x3-160x2+304x-320 
  
Edwin