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: When f(x) is divided by g(x) = x² + x - 2, we can write f(x) = q(x)·g(x) + r(x) where r(x) = 4x + 3. Since g(x) factors as (x+2)(x-1), we must have f(-2) = r(-2) and f(1) = r(1), giving us two equations to find a and b.
<p><strong>Step 1:</strong> Factor g(x) = x² + x - 2 = (x + 2)(x - 1)</p><p><strong>Step 2:</strong> By the Remainder Theorem, when f(x) is divided by g(x), we have f(x) = q(x)·g(x) + r(x), so f(x) ≡ r(x) at roots of g(x)</p><p><strong>Step 3:</strong> At x = -2: f(-2) = r(-2) = 4(-2) + 3 = -5<br/>f(-2) = 16 - 16 + 4a - 2b + 3 = 4a - 2b + 3 = -5<br/>Therefore: 4a - 2b = -8 → 2a - b = -4 ... (i)</p><p><strong>Step 4:</strong> At x = 1: f(1) = r(1) = 4(1) + 3 = 7<br/>f(1) = 1 + 2 + a + b + 3 = a + b + 6 = 7<br/>Therefore: a + b = 1 ... (ii)</p><p><strong>Step 5:</strong> Solving (i) and (ii):<br/>From (ii): b = 1 - a<br/>Substitute in (i): 2a - (1 - a) = -4 → 3a = -3 → a = -1<br/>Then b = 1 - (-1) = 2</p><p><strong>Step 6:</strong> Verify: f(x) = x⁴ + 2x³ - x² + 2x + 3<br/>Check f(-2) = 16 - 16 - 4 - 4 + 3 = -5 ✓<br/>Check f(1) = 1 + 2 - 1 + 2 + 3 = 7 ✓</p><p>∴ Answer: a = -1, b = 2 (Select options matching these values)</p>
Correct Answer: ABD