Quadratic Equations
Remainder Theorem & Polynomials
Grade 11
Question:
<p><strong>138.</strong> Let a polynomial \(P(x)\), when divided by \(x-1\), \(x-2\), \(x-3\) leaves the remainder 4, 5, 6 respectively. When \(P(x)\) is divided by \((x-1)(x-2)(x-3)\), the remainder is \(ax^2+bx+c\), then \(3a+2b+c\) is equal to:</p>
<p>(a) 3</p>
<p>(b) 4</p>
<p>(c) 5</p>
<p>(d) 6</p>
Step-by-Step Solution
Key Concept: By the Remainder Theorem, P(1)=4, P(2)=5, P(3)=6. Since the divisor is cubic, the remainder must be a quadratic ax²+bx+c, which we can determine by substituting these three conditions into the remainder polynomial.
<p><strong>Step 1:</strong> By Remainder Theorem, when P(x) is divided by (x-1)(x-2)(x-3), we have:<br/>P(x) = Q(x)·(x-1)(x-2)(x-3) + ax² + bx + c</p><p><strong>Step 2:</strong> Substitute x = 1, 2, 3 (these make the product zero):<br/>P(1) = a(1)² + b(1) + c = a + b + c = 4<br/>P(2) = a(4) + b(2) + c = 4a + 2b + c = 5<br/>P(3) = a(9) + b(3) + c = 9a + 3b + c = 6</p><p><strong>Step 3:</strong> We have the system:<br/>a + b + c = 4 ... (i)<br/>4a + 2b + c = 5 ... (ii)<br/>9a + 3b + c = 6 ... (iii)</p><p><strong>Step 4:</strong> Subtract (i) from (ii): 3a + b = 1 ... (iv)<br/>Subtract (ii) from (iii): 5a + b = 1 ... (v)</p><p><strong>Step 5:</strong> Subtract (iv) from (v): 2a = 0, so a = 0<br/>From (iv): b = 1<br/>From (i): c = 4 - 0 - 1 = 3</p><p><strong>Step 6:</strong> Therefore, 3a + 2b + c = 3(0) + 2(1) + 3 = <strong>5</strong></p><p>∴ Answer: C</p>
Correct Answer: C