Question

The polynomial \(4x^3+ax^2+bx+c\) leaves remainders 11, -3 and 41 respectively when divided by \(x+1\), \(x+2\) and \(x-2\). Then the remainder when it is divided by \(x-1\) is

• A. 19
• B. 9
• C. 14
• D. 24

Hint:

When several remainders are given, use the Remainder Theorem to create linear equations in the unknown coefficients and solve them systematically.

Answer 1

The Correct Option is B

Concept: By the Remainder Theorem, \[ P(k)=\text{remainder when }P(x)\text{ is divided by }(x-k). \] Let \[ P(x)=4x^3+ax^2+bx+c. \]

Step 1: Form equations using given remainders.
Since \(P(-1)=11\), \[ -4+a-b+c=11 \] \[ a-b+c=15 \qquad ...(1) \] Since \(P(-2)=-3\), \[ -32+4a-2b+c=-3 \] \[ 4a-2b+c=29 \qquad ...(2) \] Since \(P(2)=41\), \[ 32+4a+2b+c=41 \] \[ 4a+2b+c=9 \qquad ...(3) \]

Step 2: Solve for \(a,b,c\).
Subtracting (2) from (3), \[ 4b=-20 \] \[ b=-5 \] Substituting into (1), \[ a+c=10 \] Substituting into (2), \[ 4a+c=19 \] Subtracting, \[ 3a=9 \] \[ a=3 \] Therefore, \[ c=7 \]

Step 3: Find the remainder when divided by \(x-1\).
Required remainder \[ =P(1) \] \[ =4(1)^3+3(1)^2-5(1)+7 \] \[ =4+3-5+7 \] \[ =9 \] \centerline{{9}}