Hi Dubulyue.
This was a long one! 
There may be a better way of doing it though. 
Express
\(P(z)=z^4-z^3+z^2+2a\)
as a product of two real quadratic factors given that 1+i is a root of P(z) and a is a real number.
P(1+i)=0
\((1+i)^4-(1+i)^3+(1+i)^2+2a=0\\ (-4)-(2i-2)+(2i)+2a=0\\ (-4)-(2i-2)+(2i)+2a=0\\ -4+2+2a-2i+2i=0\\ -2+2a=0\\ a=1\)
\(P(z)=z^4-z^3+z^2+2\\ \)
I'm going to let (hope) the factors are
\((z^2+bz+c)(z^2+dz+e)\\ \begin{array}\\ &z^4\quad +&d&z^3\quad+&e&z^2\\ &&b&z^3\quad+&bd&z^2\quad+&be&z\\ &&&&c&z^2\quad+&cd&z\quad+&ce\\ &z^4&(d+b)&z^3\quad+&(e+c+bd)&z^2\quad+&(be+cd)&z\quad+&ce\\ =&z^4&(-1)&z^3\quad+&(1)&z^2\quad+&(0)&z\quad+&2 \end{array}\)
\(b+d=-1\\ \boxed{d=-1-b}\\~\\ \boxed{ce=2\quad \text{so they are either (1 and 2) or ( -1 and -2)}}\\\)
\(e+c+bd=1\\~\\ \text{If (c and e) are (2 and 1) then}\\ \qquad 3+bd=1\\ \qquad bd=-2\quad \\\ \qquad \text{so b and d must be (-1 and 2) or (1 and -2) any order}\\ or\\ \text{If (c and e) are (-2 and -1) then}\\\\ \qquad -3+bd=1\\ \qquad bd=4\quad \\\ \qquad \text{so (b and d) must be (-1 and -4) or (1 and 4) or (2,2) or (-2,-2) any order}\\\)
Now I will use the fact that d = -1 - b to check which of these could be correct.
b= -1 d=-1--1 = 0 not a pair
b=2 d=-1-2=-3 No
b= 1 d=-1 -1 = -2 so b=1 and d=-2 works
b=-2 d=-1--2 =1 so b=-2 and d=1 works
so far:
c and e are 2 and 1
and b and d are -2 and 1
\(\text{But from equating co-efficients of z}\\~\\ be+cd=0\\~\\ -2*1+2*1=0 \qquad or \qquad 1*2+ 1*-2=0\\ so\; either\\ b=1, e=2, c=1, \;\; and\;\; d=-2 \qquad (z^2+z +1)(z^2-2z +2)\\ OR \\ b=-2, e=1, c=2,\;\; and \;\; d=1 \qquad (z^2-2z +2)(z^2+z +1)\\~\\ \text{These two answers are identical}\)
\(P(z)=z^4-z^3+z^2+2=(z^2-2z +2)(z^2+z +1)\\ \)
FINITO 
+118659 
