+10 Express
P(z)=z4−z3+z2+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:
z2−2z+2
I then solved for a using P(1+i)=0 because 1+i is a root.
a=2−√2
But then after multiplying the quadratic factors and putting them equal to P(z):
(z2−2z+2)(dz2+ez+f)=z4−z3+z2+2(2−√2)
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.
+26397 Express
P(z)=z4−z3+z2+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.
Let z1=1+iIf z1 is a root of P(z) then (z−z1)=(z−(1+i)) divides P(z)If (1+i) is a root of P(z) then the complex conjugate (1−i)=z2 is a root of P(z)If z2 is a root of P(z) then (z−z2)=(z−(1−i)) divides P(z)The first real quadratic factor is (z−(1+i))⋅(z−(1−i))=[(z−1)−i]⋅[(z−1)+i]=(z−1)2−i2=(z−1)2+1=z2−2z+1+1=z2−2z+2P(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+i2)(1+2i+i2−i)+2a=0(1+2i−1)(1+2i−1−i)+2a=0(2i)(i)+2a=02i2+2a=0−2+2a=02a=2a=1also P(1−i)=0⇒a=1The second real quadratic factor is z4−z3+z2+2:(z2−2z+2)=z2+z+1so z4−z3+z2+2=(z2−2z+2)⋅(z2+z+1)
+118706 Hi Dubulyue.
This was a long one! 
There may be a better way of doing it though. 
Express
P(z)=z4−z3+z2+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=0a=1
P(z)=z4−z3+z2+2
I'm going to let (hope) the factors are
(z2+bz+c)(z2+dz+e)z4+dz3+ez2bz3+bdz2+bezcz2+cdz+cez4(d+b)z3+(e+c+bd)z2+(be+cd)z+ce=z4(−1)z3+(1)z2+(0)z+2
b+d=−1d=−1−b ce=2so they are either (1 and 2) or ( -1 and -2)
e+c+bd=1 If (c and e) are (2 and 1) then3+bd=1bd=−2 so b and d must be (-1 and 2) or (1 and -2) any orderorIf (c and e) are (-2 and -1) then−3+bd=1bd=4 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
But from equating co-efficients of z be+cd=0 −2∗1+2∗1=0or1∗2+1∗−2=0soeitherb=1,e=2,c=1,andd=−2(z2+z+1)(z2−2z+2)ORb=−2,e=1,c=2,andd=1(z2−2z+2)(z2+z+1) These two answers are identical
P(z)=z4−z3+z2+2=(z2−2z+2)(z2+z+1)
FINITO 
+26397 Express
P(z)=z4−z3+z2+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.
Let z1=1+iIf z1 is a root of P(z) then (z−z1)=(z−(1+i)) divides P(z)If (1+i) is a root of P(z) then the complex conjugate (1−i)=z2 is a root of P(z)If z2 is a root of P(z) then (z−z2)=(z−(1−i)) divides P(z)The first real quadratic factor is (z−(1+i))⋅(z−(1−i))=[(z−1)−i]⋅[(z−1)+i]=(z−1)2−i2=(z−1)2+1=z2−2z+1+1=z2−2z+2P(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+i2)(1+2i+i2−i)+2a=0(1+2i−1)(1+2i−1−i)+2a=0(2i)(i)+2a=02i2+2a=0−2+2a=02a=2a=1also P(1−i)=0⇒a=1The second real quadratic factor is z4−z3+z2+2:(z2−2z+2)=z2+z+1so z4−z3+z2+2=(z2−2z+2)⋅(z2+z+1)
+130478 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 )
