Question

If \[ ax^2+bx+c \] is exactly divisible by \((x-1)\) and \((x-2)\) but leaves a remainder \(6\) when divided by \((x+1)\), then \(a+b-c=\)

• A. \(-4\)
• B. \(-2\)
• C. \(3\)
• D. \(0\)

Hint:

If a polynomial is divisible by given linear factors, directly write it as their product times a constant.

Answer 1

The Correct Option is A

Concept: If divisible by \((x-1)\) and \((x-2)\), the polynomial must be: \[ P(x)=k(x-1)(x-2) \]

Step 1: Expand.
\[ P(x)=k(x^2-3x+2) \] Comparing: \[ a=k,\; b=-3k,\; c=2k \]

Step 2: Use remainder condition.
When divided by \((x+1)\), remainder is \(6\). By Remainder theorem: \[ P(-1)=6 \] So: \[ k(-1-1)(-1-2)=6 \] \[ k(-2)(-3)=6 \] \[ 6k=6 \] \[ k=1 \] Thus: \[ a=1,\; b=-3,\; c=2 \]

Step 3: Find \(a+b-c\).
\[ 1+(-3)-2 \] \[ =-4 \] Thus, the required answer is: \[ \boxed{-4} \]