# Thread: factorising another polynomial over C

1. ## factorising another polynomial over C

Hi, I'm stuck on this one. I don't know why though. The answer in the book is 8.
if $P(z) = z^3 + 2z^2 - 6z + a$ and $P(1 - i) = 0$, then a is equal to?
A) 4 B) -6 C) 8 D) 6 E) 5

So the first thing I saw was that there is a conjugate zero also so it looks like this
$P(z) = (z - 1 - i) (z - 1 + i) (z - z_{3})$
So then I thought i would expand out just the z's, not the z^2s etc. and let them equal the z term in the original expression, -6z
$z(-1 + i) + z(-z_{3}) + z(-1 - i) + z(-z_{3}) + z(-1 - i) + z(-1 + i) = -6z$
the z cancels out
$-1 + i - z_{3} -1 - i -z_{3} - 1 - i - 1 + i = -6$
$-4 - 2z_{3} = -6$
$-2z_{3} = -2$
$z_{3} = 1$

So then I now have (z - 1 - i)(z - 1 + i)(z - 1) and I will just expand our the constants to get the a
But this doesn't seem to work. I won't bother writing it down because I have done it about 5 times and I can't seem to get the answer. Does anyone know a method to get this one out?
David.

2. I worked it out now.
Thanks.
I put it in the form $P(z) = (z - 1 + i)(z^2 + pz + q)$ and worked out p based on the z^2 term and then worked out q based on the p and the z term, and then allowed me to work out what a is.
Thanks.