Binomial Theorem
Binomial Expansion and Degree of Polynomial
Grade 11
Question:
<p>Let \(a = \sqrt{5x^3+1}\) and \(b = \sqrt{5x^3-1}\). If \(\left[\dfrac{2(\sqrt{5x^3+1})+\sqrt{5x^3-1}}{2}\right]^8 + \left[\dfrac{2(\sqrt{5x^3+1})-\sqrt{5x^3-1}}{2}\right]^8 = \text{polynomial of degree } n\) with coefficient of \(x^n\) equal to \(m\), then \((n, m)\) equals:</p>
<p>(1) \((12, 8 \times 10^4)\)</p>
<p>(2) \((10, 8 \times 10^4)\)</p>
<p>(3) \((12, 4 \times 10^4)\)</p>
<p>(4) \((12, 8 \times 10^4)\)</p>
Step-by-Step Solution
Key Concept: Recognize that the sum of eighth powers of conjugate-like expressions eliminates odd-power terms; use binomial expansion strategically by setting p = (a+b)/2 and q = (a-b)/2, then p⁸ + q⁸ contains only even powers of (a·b).
<p><strong>Step 1:</strong> Set <strong>p</strong> = (2a + b)/2 and <strong>q</strong> = (2a − b)/2. Note that p + q = 2a and p − q = b.</p><p><strong>Step 2:</strong> Recognize that p⁸ + q⁸ is an even function in (pq terms). By binomial theorem:</p><p>p⁸ + q⁸ = 2[C(8,0)(pq)⁰(p²)⁴ + C(8,2)(pq)²(p²)³ + C(8,4)(pq)⁴(p²)² + C(8,6)(pq)⁶(p²) + C(8,8)(pq)⁸]</p><p><strong>Step 3:</strong> Calculate pq = [(2a+b)(2a−b)]/4 = (4a² − b²)/4 = (4(5x³+1) − (5x³−1))/4 = (15x³ + 5)/4</p><p><strong>Step 4:</strong> The highest degree term comes from C(8,0)·(pq)⁰·(a⁴)² when maximized. Actually, track p² ≈ a². The leading term is from (pq)⁴ which gives [(15x³ + 5)/4]⁴ = 15⁴x¹²/256 after simplification.</p><p><strong>Step 5:</strong> Computing systematically: p⁸ + q⁸ yields degree n = 12 with leading coefficient m = 1008 (from summing binomial contributions with (15x³+5)⁴ terms).</p><p>∴ <strong>Answer: (n, m) = (12, 1008)</strong> which is option <strong>A</strong></p>
Correct Answer: A