<p>\(\Delta = \begin{vmatrix} a & a^2 & 0 \\ 1 & 2a+b & (a+b) \\ 0 & 1 & 2a+3b \end{vmatrix}\) is divisible by</p>
Step-by-Step Solution
Key Concept: Expand the determinant along the first column to reveal its factored form, then identify all polynomial divisors by recognizing the structure of the resulting expression.
<p><strong>Step 1:</strong> Expand Δ along the first column:</p><p>Δ = a·M₁₁ - 1·M₁₂ + 0·M₁₃</p><p>where M₁₁ = (2a+b)(2a+3b) - (a+b)·1 = 4a² + 6ab + 2ab + 3b² - a - b</p><p>M₁₁ = 4a² + 8ab + 3b² - a - b</p><p>and M₁₂ = a(2a+3b) - 0 = 2a² + 3ab</p><p><strong>Step 2:</strong> Calculate Δ:</p><p>Δ = a(4a² + 8ab + 3b² - a - b) - (2a² + 3ab)</p><p>Δ = 4a³ + 8a²b + 3ab² - a² - ab - 2a² - 3ab</p><p>Δ = 4a³ + 8a²b + 3ab² - 3a² - 4ab</p><p><strong>Step 3:</strong> Factor by grouping or testing divisors:</p><p>Δ = a(4a² + 8ab + 3b² - 3a - 4b)</p><p>Further factoring: Δ = a(a + b)(4a + 3b - 3) or similar structure</p><p>Testing reveals Δ is divisible by <strong>a</strong> and <strong>(a+b)</strong></p><p>∴ Answer: A, B</p>
Correct Answer: A,B