<p><b>For Problems 1–3:</b> If roots of the equation \(f(x) = x^6 - 12x^5 + bx^4 + cx^3 + dx^2 + ex + 64 = 0\) are positive, then</p><p><b>Question 3:</b> Remainder when \(f(x)\) is divided by \(x - 1\) is</p>
Step-by-Step Solution
Key Concept: By Vieta's formulas for a degree-6 polynomial, the product of all 6 roots equals 64 (the constant term). If all roots are positive and their product is 64, we can use the constraint that the coefficient of x^5 is -12 (sum of roots). The remainder when f(x) is divided by (x-1) is simply f(1).
<p><strong>Step 1:</strong> Apply the Remainder Theorem. When f(x) is divided by (x−1), the remainder is f(1).</p><p><strong>Step 2:</strong> Calculate f(1):</p><p>f(1) = 1 − 12(1) + b + c + d + e + 64</p><p>f(1) = 1 − 12 + b + c + d + e + 64 = 53 + b + c + d + e</p><p><strong>Step 3:</strong> From Vieta's formulas: product of roots = 64, and sum of roots = 12. For positive roots with product 64 and sum 12, by AM-GM inequality: 12/6 ≥ ⁶√64, which gives 2 ≥ 2. Equality holds when all roots equal 2.</p><p><strong>Step 4:</strong> If all 6 roots equal 2, then f(x) = (x−2)⁶. Expanding: f(x) = x⁶ − 12x⁵ + 60x⁴ − 160x³ + 240x² − 192x + 64</p><p>Therefore: b = 60, c = −160, d = 240, e = −192</p><p><strong>Step 5:</strong> f(1) = 1 − 12 + 60 − 160 + 240 − 192 + 64 = <strong>1</strong></p><p>∴ Answer: A</p>
Correct Answer: A