Thread: Finding Linear Factors Over C

1. Finding Linear Factors Over C

Hi, the problem I have is the following:

z^2 + 4z + 10, finding the factors over C.

I have gotten up to: (-2 + (sqrt6)i) and (-2 - (sqrt6)i).

I also know, however, that the answers should be

(z + 2 + (sqrt6)i)(z + 2 - (sqrt6)i).

The problem is i can't see how i can change my original answer to this one. What do I need to do?

Thanks

2. Originally Posted by Stevo_Evo_22
Hi, the problem I have is the following:

z^2 + 4z + 10, finding the factors over C.

I have gotten up to: (-2 + (sqrt6)i) and (-2 - (sqrt6)i).

I also know, however, that the answers should be

(z + 2 + (sqrt6)i)(z + 2 - (sqrt6)i).

The problem is i can't see how i can change my original answer to this one. What do I need to do?

Thanks
$\displaystyle z^2+4z+10,~~F_1,F_2\in\mathbb{C}$

(where $\displaystyle F_1$ and $\displaystyle F_2$ are complex factors of $\displaystyle z$)

So when you solved for z, I'm sure you saw that $\displaystyle z=\frac{-4\pm\sqrt{16-40}}{2}\implies z=-2\pm\sqrt{6}i$

note that these are zeros.

However, to get factors, we want to get the expressions $\displaystyle F_1=0$ and $\displaystyle F_2=0$, so we see that $\displaystyle z+2-\sqrt{6}i=0$ and $\displaystyle z+2+\sqrt{6}i=0$

So we see that $\displaystyle z^2+4z+10=(z+2+\sqrt{6}i)(z+2-\sqrt{6}i)$

I hope this is clear enough. If you have any additional questions, feel free to ask!

--Chris

3. Originally Posted by Stevo_Evo_22
Hi, the problem I have is the following:

z^2 + 4z + 10, finding the factors over C.

I have gotten up to: (-2 + (sqrt6)i) and (-2 - (sqrt6)i).

I also know, however, that the answers should be

(z + 2 + (sqrt6)i)(z + 2 - (sqrt6)i).

The problem is i can't see how i can change my original answer to this one. What do I need to do?

Thanks
$\displaystyle z^2+4z+10$

you got $\displaystyle (-2- i \sqrt{6}) \;and\; (-2+ i \sqrt{6})$

Now the factors are

$\displaystyle [z-(-2- i \sqrt{6})].[z-(-2+ i \sqrt{6})]$

$\displaystyle [z+2 + i \sqrt{6})].[z+2- i \sqrt{6})]$

4. Thank you very much. At the start I wasn't sure exactly what i was finding, so knowing that they were possible solutions of z, it's cleared it up a lot