<p>Let \(1, \omega, \omega^2\) be the cube roots of unity. The least possible degree of a polynomial with real coefficients having roots \(2\omega, (2+3\omega), (2+3\omega^2), (2-\omega-\omega^2)\) is ___.</p>
Step-by-Step Solution
Key Concept: Since ω and ω² are complex conjugates (both non-real cube roots of unity), any polynomial with real coefficients must have conjugate pairs. Use the fact that 1 + ω + ω² = 0 to simplify roots and identify conjugate pairs.
<p><strong>Step 1:</strong> Identify the relationship between roots using ω and ω². Since ω is a primitive cube root of unity: ω³ = 1, ω² = ω̄ (complex conjugate), and 1 + ω + ω² = 0.</p><p><strong>Step 2:</strong> Simplify root (2−ω−ω²). Using 1 + ω + ω² = 0, we get ω + ω² = −1, so 2 − ω − ω² = 2 − (−1) = 3.</p><p><strong>Step 3:</strong> Identify conjugate pairs. Since ω̄ = ω²:</p><ul><li>2ω and 2ω² are complex conjugates</li><li>(2+3ω) and (2+3ω²) are complex conjugates</li><li>3 is real</li></ul><p><strong>Step 4:</strong> Count minimum roots for real coefficients. For a polynomial with real coefficients:</p><ul><li>The conjugate pair {2ω, 2ω²} requires degree ≥ 2</li><li>The conjugate pair {(2+3ω), (2+3ω²)} requires degree ≥ 2</li><li>The real root 3 requires degree ≥ 1</li></ul><p><strong>Step 5:</strong> The minimum degree polynomial must include all conjugate pairs: one polynomial of degree 2 from {2ω, 2ω²}, one polynomial of degree 2 from {(2+3ω), (2+3ω²)}, and linear factor (x−3).</p><p>∴ Answer: <strong>5</strong></p>
Correct Answer: 5