# Finding real and complex factors of polynoms

1. Sep 4, 2011

### Noorac

Hi. I'm having some trouble factoring complex polynomials. I'm sure it's very easy assignments, but I'm missing some key-information as to how to solve these tasks.

The task is to find the complex and real factors of the polynomials(since I'm not that articulate in english I will provide the answer for the first task, just so it's easier to see what type of answers the task is looking for)

a) z3 + 8

b) z4 - 1

c) I will provide this one if I still don't get it.

(I decided to list two tasks since I find that it is easier to compare the solution to different tasks than to just try and understand how one solution works. However, at this point I'm happy with any help I can get, since I'm pretty much stuck).

a) (z + 2)(z - 1 - i$\sqrt{3}$)(z - 1 + i$\sqrt{3}$) = (z + 2)(z2 - 2z + 4)

(Side note: i = $\sqrt{-1}$)

PS: telling me how to aproach the task and solve it is just as good(if not better), than just solving the task. Then I can actually try to solve it myself.

Thanks, Noorac=)

2. Sep 4, 2011

### wukunlin

umm... I'm not sure why you (or they) want to factorize these polynomials, persoanlly I find it easiest to do the following: (I'll use a) as an example)

1) make $$z^3=-8$$
2) you will have the most obvious solution, ie z = -2 or z+2=0
3) use long division to perform $$\frac{z^3}{z+2}$$ and you should obtain the other factor

another method is to draw the most obvious solution on an argand plane. The amount of solutions of the polynomial is equal to the highest power of z, and all solutions have the same modulus and same angular spacing to each other, you can then work out the complex roots that way

3. Sep 4, 2011

### Noorac

Wukunlin:

(The reason to finding the factors are non other than practice, it's a task taken from my textbook).

I've tried a), and as far as I know, made it, yet I can't make it for the other ones(b).
Heres how i solved a)

z3 + 8 => $\sqrt[3]{8}$ei*$\frac{pi}{3}$ = 2(cos$\frac{pi}{3}$ + i*sin$\frac{pi}{3}$) = 1 + i$\sqrt{3}$

Wich has gotten me a bit closer to the solution(the complex root). The answer is on the form:

Cn(z - r1)(z - r2)...
And since this is a real polynomial, the conjugent(is this the right english word for switching the sign infront of the complex part of the equation?) is also a root, meaning I have:

Cn(z - (1 + i$\sqrt{3}$))(z - (1 - i$\sqrt{3}$)). Now, all I need to find out is what Cn is. How?

For the real factor, I just decided to take the third root of z3 + 8 = z + 2
and perform long division(I'll spare the actual division):

z3 + 8 : z + 2 = z2 - 2z + 4

This gives me the correct real factor: (z + 2)(z2 - 2z + 4)

Is this the correct way to solve this? Or is there a more elegant way? If it is correct, I still have trouble doing b)

Edit; spelling errors

4. Sep 4, 2011

### wukunlin

$C_n$ I believe corresponds to the coefficient of z in this case (1)

you pretty much did what I meant to say, except in your first line, you could have set the argument to 0 instead of $\frac{\pi}{3}$

then when you do the long division you will get $(z^2 - 2z + 4)$, and then you could find the other 2 roots using the quadratic formula

5. Sep 4, 2011

### lanedance

conjugate, is the word

and its worth remembering any complex roots will always come in conjugate pairs as well

i find a good way is to write it is as
$$z^3+8 = 0$$

then re-arrange and write in complex polar form $z=re^{i \theta}$
$$z^3=r^3e^{i 3 \theta}=-8$$

from there, you should be able to find the thetas that only give real values in the exponential
$$3 \theta = n \pi$$
$$\theta = n \frac{\pi}{3}$$
$$n =0,1,2$$

and the corresponding r should follow