<p><strong>913.</strong> Let \(f\) be monic cubic polynomial such that \(f(1)=1^4-1\), \(f(2)=2^4-2\) and \(f(3)=3^4-3\). If \(f(4)=N\), then find the number of prime factors of \(N\).</p>
Step-by-Step Solution
Key Concept: Define g(x) = f(x) - (x⁴ - x). Since f is cubic and x⁴ - x is quartic, g(x) is a quartic polynomial with roots at x = 1, 2, 3. Use the leading coefficient constraint to find f(4).
<p><strong>Step 1:</strong> Let g(x) = f(x) - (x⁴ - x). We know g(1) = g(2) = g(3) = 0.</p><p><strong>Step 2:</strong> Since f is monic cubic, f(x) = x³ + ax² + bx + c. Then g(x) = x³ + ax² + bx + c - x⁴ + x = -x⁴ + x³ + ax² + bx + (c + x).</p><p><strong>Step 3:</strong> More precisely, g(x) = f(x) - (x⁴ - x) has degree 4 with leading coefficient -1 (from -x⁴). Since g(x) has roots at x = 1, 2, 3, we write: g(x) = -(x - 1)(x - 2)(x - 3)(x - r) for some root r.</p><p><strong>Step 4:</strong> Expanding: g(x) = -(x - 1)(x - 2)(x - 3)(x - r). The coefficient of x⁴ is -1 ✓. Since f is monic cubic, comparing f(x) = g(x) + x⁴ - x gives us the leading term. The coefficient of x³ in g(x) must match: -(−1−2−3−r) = 6 + r from expansion, which equals 1 (coefficient of x³ in f). So r = -5.</p><p><strong>Step 5:</strong> Therefore g(x) = -(x - 1)(x - 2)(x - 3)(x + 5).</p><p><strong>Step 6:</strong> f(4) = g(4) + (4⁴ - 4) = -(3)(2)(1)(9) + (256 - 4) = -54 + 252 = 198.</p><p><strong>Step 7:</strong> 198 = 2 × 99 = 2 × 9 × 11 = 2 × 3² × 11.</p><p>∴ Answer: <strong>4 prime factors</strong> (counting 2, 3, 3, 11)</p>
Correct Answer: 4