Complex Numbers
Cube Roots of Unity
Grade 11
Question:
<p>If <span>f(x) = g(x^3) + xh(x^3)</span> is divisible by <span>x^2 + x + 1</span>, then</p>
<p>(a) <span>g(x)</span> is divisible by <span>(x - 1)</span> but not <span>h(x)</span></p>
<p>(b) <span>h(x)</span> is divisible by <span>(x - 1)</span> but not <span>g(x)</span></p>
<p>(c) both <span>g(x)</span> and <span>h(x)</span> are divisible by <span>(x - 1)</span></p>
<p>(d) None of the above</p>
Step-by-Step Solution
Key Concept: If f(x) is divisible by x² + x + 1, then f(ω) = 0 and f(ω²) = 0, where ω is a complex cube root of unity. Use these conditions to establish constraints on g(x) and h(x).
<p><strong>Step 1: Identify roots of the divisor.</strong></p><p>Let ω be a primitive cube root of unity, so ω³ = 1 and ω ≠ 1. The roots of x² + x + 1 = 0 are ω and ω², where ω² + ω + 1 = 0.</p><p><strong>Step 2: Apply divisibility condition at x = ω.</strong></p><p>Since f(x) is divisible by x² + x + 1, we have f(ω) = 0:</p><p>f(ω) = g(ω³) + ω·h(ω³) = g(1) + ω·h(1) = 0</p><p>Therefore: g(1) + ω·h(1) = 0 ... (i)</p><p><strong>Step 3: Apply divisibility condition at x = ω².</strong></p><p>Similarly, f(ω²) = 0:</p><p>f(ω²) = g((ω²)³) + ω²·h((ω²)³) = g(ω⁶) + ω²·h(ω⁶) = g(1) + ω²·h(1) = 0</p><p>Therefore: g(1) + ω²·h(1) = 0 ... (ii)</p><p><strong>Step 4: Solve the system of equations.</strong></p><p>From equations (i) and (ii):</p><p>g(1) + ω·h(1) = 0</p><p>g(1) + ω²·h(1) = 0</p><p>Subtracting: (ω - ω²)·h(1) = 0</p><p>Since ω ≠ ω² (they are distinct roots), we have h(1) = 0.</p><p><strong>Step 5: Find g(1).</strong></p><p>Substituting h(1) = 0 into equation (i):</p><p>g(1) + ω·(0) = 0</p><p>Therefore: g(1) = 0</p><p><strong>Step 6: Conclusion about divisibility by (x - 1).</strong></p><p>Since g(1) = 0, the polynomial g(x) is divisible by (x - 1).</p><p>Since h(1) = 0, the polynomial h(x) is divisible by (x - 1).</p><p><strong>∴ Answer: C</strong> Both g(x) and h(x) are divisible by (x - 1).</p>
Correct Answer: C