Find the irreducible quadratic factors of

[bingo92]

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

The Attempt at a Solution

Im stumped...this is all I've got:

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

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


Any guidance is greatly appreciated!

 

[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 [itex]\alpha=rcis(\theta)[/itex] and its conjugate [itex]\overline{\alpha}=rcis(-\theta)[/itex] then

[tex](z-\alpha)(z-\overline{\alpha})=(z^2-(\alpha+\overline{\alpha})z+\alpha\overline{\alpha})[/tex]

and

[tex]\alpha+\overline{\alpha}=2rcos\theta[/tex]

[tex]\alpha\overline{\alpha}=r^2[/tex]

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.

 

[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.

 

[Mentallic]

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.

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