+10 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.
I managed to find one quadratic factor by using the complex root and its conjugate:
\(z^2-2z+2\)
I then solved for a using P(1+i)=0 because 1+i is a root.
\(a = 2-\sqrt2\)
But then after multiplying the quadratic factors and putting them equal to P(z):
\((z^2-2z+2)(dz^2+ez+f)=z^4-z^3+z^2+2(2-\sqrt2)\)
I can't find a value for d, e, and f which they all agree on.
I might have gone wrong at any point during my working, so don't assume what I've written is right. Any help figuring out what went wrong would be appreciated.
+26388 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.
\(\begin{array}{|lcll|} \hline \text{Let } z_1 = 1+i\\\\ \text{If } z_1 \text{ is a root of } P(z) \text{ then } (z-z_1)=(z-(1+i)) \text{ divides } P(z) \\ \text{If } (1+i) \text{ is a root of } P(z) \text{ then the complex conjugate } (1-i)=z_2 \text{ is a root of } P(z) \\ \text{If } z_2 \text{ is a root of } P(z) \text{ then } (z-z_2)=(z-(1-i)) \text{ divides } P(z) \\ \text{The first real quadratic factor is }\\ (z-(1+i))\cdot (z-(1-i)) \\ = [(z-1)-i]\cdot [(z-1)+i]\\ = (z-1)^2 -i^2 \\ = (z-1)^2 +1 \\ = z^2-2z+1+1 \\ = z^2-2z+2 \\\\ P(1+i)=0\\ (1+i)^4-(1+i)^3+(1+i)^2+2a = 0 \\ (1+i)^2[ (1+i)^2-(1+i)+1]+2a = 0 \\ (1+i)^2[ (1+i)^2-i]+2a = 0 \\ (1+2i+i^2)( 1+2i+i^2 -i)+2a = 0 \\ (1+2i-1)( 1+2i-1-i )+2a = 0 \\ ( 2i )( i )+2a = 0 \\ 2i^2 +2a = 0 \\ -2 +2a = 0 \\ 2a = 2 \\ a = 1 \\\\ \text{also } P(1-i)=0 \Rightarrow a = 1 \\\\ \text{The second real quadratic factor is } z^4-z^3+z^2+2 : (z^2-2z+2) = z^2+z+1 \\ \text{so } z^4-z^3+z^2+2 = (z^2-2z+2)\cdot ( z^2+z+1 ) \\ \hline \end{array}\)
+118658 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 
+26388 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.
\(\begin{array}{|lcll|} \hline \text{Let } z_1 = 1+i\\\\ \text{If } z_1 \text{ is a root of } P(z) \text{ then } (z-z_1)=(z-(1+i)) \text{ divides } P(z) \\ \text{If } (1+i) \text{ is a root of } P(z) \text{ then the complex conjugate } (1-i)=z_2 \text{ is a root of } P(z) \\ \text{If } z_2 \text{ is a root of } P(z) \text{ then } (z-z_2)=(z-(1-i)) \text{ divides } P(z) \\ \text{The first real quadratic factor is }\\ (z-(1+i))\cdot (z-(1-i)) \\ = [(z-1)-i]\cdot [(z-1)+i]\\ = (z-1)^2 -i^2 \\ = (z-1)^2 +1 \\ = z^2-2z+1+1 \\ = z^2-2z+2 \\\\ P(1+i)=0\\ (1+i)^4-(1+i)^3+(1+i)^2+2a = 0 \\ (1+i)^2[ (1+i)^2-(1+i)+1]+2a = 0 \\ (1+i)^2[ (1+i)^2-i]+2a = 0 \\ (1+2i+i^2)( 1+2i+i^2 -i)+2a = 0 \\ (1+2i-1)( 1+2i-1-i )+2a = 0 \\ ( 2i )( i )+2a = 0 \\ 2i^2 +2a = 0 \\ -2 +2a = 0 \\ 2a = 2 \\ a = 1 \\\\ \text{also } P(1-i)=0 \Rightarrow a = 1 \\\\ \text{The second real quadratic factor is } z^4-z^3+z^2+2 : (z^2-2z+2) = z^2+z+1 \\ \text{so } z^4-z^3+z^2+2 = (z^2-2z+2)\cdot ( z^2+z+1 ) \\ \hline \end{array}\)
+129839 Going from this point in Melody's answer....
P(z) = z^4 - z^3 + z^2+ 2
We know that 1 + i is a root....therefore (1 - i) must also be a root.....thus, the resulting polynomial is
[z - ( 1 + i ) ] [ z - (1 - i ) ] =
z^2 -z(1 + i) -z(1-i) + 2 =
z^2 - 2z + 2
Now.....using some polynomial long division, we can determine the other factor
z^2 + z + 1
z^2 - 2z + 2 [ z^4 - z^3 + z^2 + 2 ]
z^4 - 2z^3 + 2z^2
________________________
z^3 - z^2 + 2
z^3 -2z^2 + 2z
-------------------------------
z^2 - 2z + 2
z^2 -2z + 2
_____________
And the two factors are ( z^2 - 2z + 2 ) ( z^2 + z + 1 )
