Thread: Factorising this polynomial over C

1. Factorising this polynomial over C

Hi there

I'm having trouble getting this problem out.
I have to factorise this over c.
$\displaystyle 4z^2 - 4z + 17$
$\displaystyle z^2 - z + \frac{17}{4}$
$\displaystyle (z^2 - z + \frac{1}{4}) + \frac{17}{4} - \frac{1}{4}$
$\displaystyle (z - \frac{1}{2})^2 + \frac{16}{4}$
$\displaystyle (z - \frac{1}{2})^2 + 4$
$\displaystyle (z - \frac{1}{2})^2 - (2i)^2$
$\displaystyle (z - \frac{1}{2} - 2i)(z - \frac{1}{2} + 2i)$

However, the answer in the book is this:
$\displaystyle (2z - 1 + 4i)(2z - 1 - 4i)$
Is this the same as what I had?
Also, have I put this complex number question in the right forum? What section should a question like this go?

2. Originally Posted by Bucephalus
Hi there

I'm having trouble getting this problem out.
I have to factorise this over c.
$\displaystyle 4z^2 - 4z + 17$
$\displaystyle z^2 - z + \frac{17}{4}$
$\displaystyle (z^2 - z + \frac{1}{4}) + \frac{17}{4} - \frac{1}{4}$
$\displaystyle (z - \frac{1}{2})^2 + \frac{16}{4}$
$\displaystyle (z - \frac{1}{2})^2 + 4$
$\displaystyle (z - \frac{1}{2})^2 - (2i)^2$
$\displaystyle (z - \frac{1}{2} - 2i)(z - \frac{1}{2} + 2i)$

However, the answer in the book is this:
$\displaystyle (2z - 1 + 4i)(2z - 1 - 4i)$
Is this the same as what I had?
Also, have I put this complex number question in the right forum? What section should a question like this go?
You lost a $\displaystyle 4$ in your very first step

$\displaystyle \displaystyle 4\left(z^2 - z + \frac{17}{4} \right)$

Now in your final answer if you break $\displaystyle 4=2\cdot 2$

and put it back into each factor your answers will be the same.

3. Originally Posted by Bucephalus
Hi there

I'm having trouble getting this problem out.
I have to factorise this over c.
$\displaystyle 4z^2 - 4z + 17$
$\displaystyle z^2 - z + \frac{17}{4}$
$\displaystyle (z^2 - z + \frac{1}{4}) + \frac{17}{4} - \frac{1}{4}$
$\displaystyle (z - \frac{1}{2})^2 + \frac{16}{4}$
$\displaystyle (z - \frac{1}{2})^2 + 4$
$\displaystyle (z - \frac{1}{2})^2 - (2i)^2$
$\displaystyle (z - \frac{1}{2} - 2i)(z - \frac{1}{2} + 2i)$

However, the answer in the book is this:
$\displaystyle (2z - 1 + 4i)(2z - 1 - 4i)$
Is this the same as what I had?
Also, have I put this complex number question in the right forum? What section should a question like this go?
Your coefficient of z^2 is 1. The coefficient in the question is 4.

I suggest the following:

$\displaystyle 4z^2 - 4z + 17 = 4(z^2 -z) + 17 = 4\left(\left[z - \frac{1}{2}\right]^2 - \frac{1}{4}\right) + 17 = (2z - 1)^2 + 16 = ....$