# Math Help - Factorising a polynomail over c

1. ## Factorising a polynomail over c

If z - 3i is a factor of $P(z) = 2z^4 - 4z^3 + 21z^2 - 36z + 27$, find the remaining factors.

So I did this:
Step 1: if z - 3i is a factor, then z + 3i is a factor.
Step 2: $P(z) = (z - z_{1})(z - z_{2})(z - z_{3})(z - z_{4})$
Step 3: $P(z) = (z - z_{1})(z - z_{2})(2z^2 + pz +q)$
Step 4: $2z^4 - 4z^3 + 21z^2 - 36z + 27 = (z - 3i)(z + 3i)(2z^2 + pz +q)$
Step 5: $2z^4 - 4z^3 + 21z^2 - 36z + 27 = (z^2 + 9)(2z^2 + pz + q)$
Step 6: $9q = 27$
Step 7: $q = 3$
Step 8: $9p = -36$
Step 9: $p = -4$
Step 10: Find the factors from the quadratic. $2z^2 - 4z + 3$
Step 11: $2(z^2 - 2z + \frac{3}{2})$
Step 12: $2[(z^2 - 2z + 1) + \frac{3}{2} - \frac{2}{2})]$
Step 13: $2[(z - 1)^2 + \frac{1}{2})]$
Step 14: $2[(z - 1)^2 - (\frac{i}{\sqrt{2}})^2)]$
Step 15: $2[(z - 1 + \frac{i}{\sqrt{2}})(z - 1 - \frac{i}{\sqrt{2}})]$
So this is the answer I got, but the book says just that these are the two remaining factors without the leading 2, as follows:
(z - 1 + \frac{i}{\sqrt{2}})(z - 1 - \frac{i}{\sqrt{2}}

So did I get it wrong or is the 2 not really classified as part of the factor for some reason? I can't understand why the 2 isn't included in the answer in the back of the book.

David.

2. I think it's because 2 doesn't qualify as a root.

3. Your working is absolutely fine. The 2 is omitted as it's a constant term. I've struggled to explain why for a few minutes, with absolutely no success, however. I'll think about it, or perhaps else will help whilst I try to find the correct wording.

Edit: No need, courtesy of VonNemo

4. $\dfrac{z^4-4z^3+21z^2-36z+27}{z^2+9}=2z^2-4z+3$.
Now find two roots of $2z^2-4z+3$.

5. Thanks.
I will take it that I factorised it correctly then.
David.