# Find the irreducible quadratic factors of

1. Dec 14, 2011

### bingo92

find the irreducible quadratic factors of z^(4)+4

3. The attempt at a solution

Im stumped...this is all ive got:

[(z^(2))^2]-[(2i)^2]

(z^(2)-2i)(z^2+2i)

Any guidance is greatly appreciated!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 14, 2011

### Mentallic

So you've already noticed that a difference of two squares won't work. The way to answer this is a little more complicated.

Can you find all the complex roots of that polynomial?

Also, notice that if we have a complex root of the form $\alpha=rcis(\theta)$ and its conjugate $\overline{\alpha}=rcis(-\theta)$ then

$$(z-\alpha)(z-\overline{\alpha})=(z^2-(\alpha+\overline{\alpha})z+\alpha\overline{\alpha})$$

and

$$\alpha+\overline{\alpha}=2rcos\theta$$

$$\alpha\overline{\alpha}=r^2$$

which are both real, which tells us if two roots are conjugates of each other then the quadratic that has those roots is real and obviously irreducible.

3. Dec 14, 2011

### eumyang

I tried equating coefficients. I assume that the factorization will be in the form of
(z2 + az + b)(z2 + cz + d).
Multiply the trinomials and you'll have a 5-term polynomial:
z4 + ?z3 + ?z2 + ?z + bd
(I'll let you fill in the "?").
Let this equal z4 + 4. This means that the coefficients for z3, z2 and z must be zero. You'll end up with 4 equations and 4 unknowns. Since solving this particular system is daunting, you could make some assumptions as to what b and d are. If you let b = d = 2, the system reduces quite nicely, and you'll find a and c easily enough.

I wouldn't say that this method can be used to factor any quartic in the form of z4 + c to irreducible quadratics. I only tried it because the constant term in the quartic is small.

4. Dec 14, 2011

### Mentallic

While that way works too, I'm pretty sure the expected method is the one that I pointed towards.