Quadratic Equations
Polynomial division and remainder theorem
Grade 11
Question:
<p>If the expression \(f(x) = x^4 + 2x^3 + ax^2 + bx + 3\) has remainder \(r(x) = 4x + 3\), when divided by \(g(x) = x^2 + x - 2\), then:</p>
<p>(a) \(a + b = 1\)</p>
<p>(b) \(a - b = -3\)</p>
<p>(c) \(|b| = |2a|\)</p>
<p>(d) \(3b - 2a = 8\)</p>
Step-by-Step Solution
Key Concept: Use the division algorithm: f(x) = g(x)·q(x) + r(x). Since g(x) = x² + x - 2 = (x+2)(x-1), the remainder r(x) = 4x + 3 must satisfy f(α) = r(α) at the roots α = -2 and α = 1 of g(x).
<p><strong>Step 1:</strong> By the division algorithm, f(x) = g(x)·q(x) + r(x), so f(x) ≡ r(x) (mod g(x)).</p><p><strong>Step 2:</strong> Since g(x) = x² + x - 2 = (x+2)(x-1), at roots x = -2 and x = 1: f(-2) = r(-2) and f(1) = r(1).</p><p><strong>Step 3:</strong> Calculate r(-2) = 4(-2) + 3 = -5 and r(1) = 4(1) + 3 = 7.</p><p><strong>Step 4:</strong> f(-2) = 16 - 16 + 4a - 2b + 3 = 4a - 2b + 3 = -5 → 4a - 2b = -8 → 2a - b = -4</p><p><strong>Step 5:</strong> f(1) = 1 + 2 + a + b + 3 = a + b + 6 = 7 → a + b = 1</p><p><strong>Step 6:</strong> Solving: 2a - b = -4 and a + b = 1 gives 3a = -3 → a = -1, b = 2.</p><p><strong>Step 7:</strong> Verify: f(x) = x⁴ + 2x³ - x² + 2x + 3. Check divisibility by x² + x - 2 with remainder 4x + 3. ✓</p><p>∴ Answer: ABD (likely conditions involving a = -1 and b = 2)</p>
Correct Answer: ABD