Quadratic Equations
Polynomial remainder theorem
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 dividing P(x) by the cubic (x-1)(x-2)(x-3) gives remainder ax²+bx+c (degree < 3), we have P(x) = Q(x)(x-1)(x-2)(x-3) + ax²+bx+c. Evaluating at x=1,2,3 gives three linear equations in a,b,c.
<p><strong>Step 1:</strong> By Remainder Theorem, when P(x) is divided by (x-1)(x-2)(x-3), we can write:</p><p>P(x) = Q(x)(x-1)(x-2)(x-3) + ax²+bx+c</p><p><strong>Step 2:</strong> Substitute x=1: P(1) = 0 + a(1)²+b(1)+c = a+b+c = 4</p><p><strong>Step 3:</strong> Substitute x=2: P(2) = 0 + a(4)+b(2)+c = 4a+2b+c = 5</p><p><strong>Step 4:</strong> Substitute x=3: P(3) = 0 + a(9)+b(3)+c = 9a+3b+c = 6</p><p><strong>Step 5:</strong> Solve the system:</p><p>Equation (2) - Equation (1): 3a+b = 1</p><p>Equation (3) - Equation (2): 5a+b = 1</p><p>Subtracting: 2a = 0 ⟹ a = 0</p><p>From 3a+b = 1: b = 1</p><p>From a+b+c = 4: 0+1+c = 4 ⟹ c = 3</p><p><strong>Step 6:</strong> Calculate 3a+2b+c = 3(0)+2(1)+3 = 0+2+3 = 5</p><p>∴ Answer: C (assuming answer choice is 5)</p>
Correct Answer: C