<p>Let \(f(x) = x^2 + bx + c\), where \(b, c \in \mathbb{R}\). If \(f(x)\) is a factor of both \(x^4 + 6x^2 + 25\) and \(3x^4 + 4x^2 + 28x + 5\), then the least value of \(f(x)\) is</p>
Step-by-Step Solution
Key Concept: If a quadratic f(x) divides two polynomials, it must divide any linear combination of them. Use polynomial division and the constraint that f(x) is a common factor to determine b and c uniquely.
<p><strong>Step 1:</strong> Since f(x) = x² + bx + c divides both P(x) = x⁴ + 6x² + 25 and Q(x) = 3x⁴ + 4x² + 28x + 5, it must also divide their linear combinations.</p><p><strong>Step 2:</strong> Compute 3P(x) - Q(x) = 3(x⁴ + 6x² + 25) - (3x⁴ + 4x² + 28x + 5) = 14x² - 28x + 70 = 14(x² - 2x + 5).</p><p><strong>Step 3:</strong> Since f(x) divides both P(x) and the linear combination above, f(x) must divide 14(x² - 2x + 5). Since f(x) is monic quadratic, we must have f(x) = x² - 2x + 5.</p><p><strong>Step 4:</strong> Verify: With b = -2, c = 5, compute the minimum value of f(x) = x² - 2x + 5 using completing the square: f(x) = (x - 1)² + 4. The minimum occurs at x = 1, giving f(1) = 4.</p><p><strong>Step 5:</strong> Confirm x² - 2x + 5 divides both original polynomials by polynomial division (both divisions yield exact results with no remainder).</p><p>∴ Answer: D (minimum value = 4)</p>
Correct Answer: D