What is the least possible degree of polynomial with real coefficient having 2ω², 3 + 4ω, 3 + 4ω², 5 - ω - ω² as roots

First of all, we should not think that because of the 4 roots, the degree of the polynomial will be 4.

We need to remember that complex roots always occur in pairs.

Now,

\(ω = {\cos \frac{2\pi}{3}}+{i \sin \frac{2\pi}{3}}\)

\(= - \frac{1}{2} + \frac{√ 3}{2} i\)

\(ω^2 = \cos \frac{4 \pi}{3} + i \sin \frac{4 \pi}{3}\)

\(= - \frac{1}{2} - \frac{√ 3}{2} i\)

Now, to find the value of the given pair roots,

1) 2ω² = -1 - √3 i

∴ -1 + √3i will also be a root.

2) 3 + 4ω = 3 - 2 + 2√3i

= 1 + 2√3i

∴ 1 - 2√3i will also be a root.

3) 3 + 4ω² = 3 - 2 - 2√3i

= 1 - 2√3i

i.e, 3 + 4ω and 3 + 4ω² are conjugate pairs.

\(4) ~5 - \omega - \omega^2 = 5 + \frac{1}{2} - \frac{\sqrt 3}{2}i + \frac{1}{2} + \frac{\sqrt 3}{2}i\)

= 6

i.e. 6 is a root.

Total we have 5 roots.

The minimum degree of the polynomial is 5.