Question

If a polynomial in \(x\) leaves remainders \(-2,19\) respectively when divided by \(x+3\) and \(x-4\), then the remainder it leaves when it is divided by \[ x^2-x-12 \] is

• A. \(2x+5\)
• B. \(3x+7\)
• C. \(3x-7\)
• D. \(5x-2\)

Hint:

When dividing by a quadratic, always assume the remainder as \(ax+b\).

Answer 1

The Correct Option is B

Concept: If dividing by a quadratic, the remainder must be of degree less than 2. So let the remainder be: \[ R(x)=ax+b \] Factorize: \[ x^2-x-12=(x+3)(x-4) \]

Step 1: Use the given remainders.
By Remainder Theorem: When divided by \(x+3\), \[ P(-3)=-2 \] So: \[ R(-3)=-2 \] \[ -3a+b=-2 \qquad ...(1) \] When divided by \(x-4\), \[ P(4)=19 \] So: \[ R(4)=19 \] \[ 4a+b=19 \qquad ...(2) \]

Step 2: Solve the equations.
Subtract (1) from (2): \[ 7a=21 \] \[ a=3 \] Substitute into (2): \[ 4(3)+b=19 \] \[ 12+b=19 \] \[ b=7 \]

Step 3: Write the remainder.
\[ R(x)=3x+7 \] Thus, the required answer is: \[ \boxed{3x+7} \]