IPMAT/Algebra•27d
bhavaydhillon

Tried everywhere, fitted nowhere.

I've asked this question in many places and got several solutions and methods, but I still can't get a fundamental answer. I already know the fundamental truths to solve this, so just explaining the fundamentals won't help. I understand the method well; many try to give you steps for this specific type of question, but that only treats the symptom, not the problem.

**Someone says Take that road to Karachi, the reason is also stated: it doesn't have pits and traffic. **

YOU TELL ME THE ROAD; I REACH KARACHI ONCE. YOU TELL ME HOW TO PICK A ROAD; I REACH ANY CITY, AH... question, I mean).

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Comments (1)

M
mayank_mamnani
•27d
Correct

Here is the direct shortcut:

  1. Set the remainder form: Since the divisor (x-1)(x-2) has power 2, your remainder is just ax + b.(Note: If the divisor is cubic, the remainder becomes quadratic(degree-1) like ax^2 + bx + c, but that’s not really likely for IPMAT).
  2. **Use the given remainders directly:**Divided by (x - 1), remainder is 2 => put x = 1 => a(1) + b = 2
  3. Divided by (x - 2) remainder is 1 => put x = 2 => a(2) + b = 1

Solve for a and B:

a = -1

b = 3.

Answer: -x + 3 (or 3 - x).

The main logic: Whenever you see a question like this, set remainder = ax + b, put the roots (x = 1, 2) equal to the given remainder values, and solve for a and b. That's it.

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