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December 19, 2018

2016 ISI Objective Solution Problem 1


The polynomial \(x^7+x^2+1\) is divisible by

    • (A) \(x^5-x^4+x^2-x+1\)             (B) \(x^5-x^4+x^2+1\)
    • (C)   \(x^5+x^4+x^2+x+1\)          (D)   \(x^5-x^4+x^2+x+1\)

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Understanding the Problem:

The problem is easy to understand right?


Before scrolling down, your first task is to try the problem YOURSELF.

All the best

We will guide you along a short path to the solution in a step by step approach.

Hint 1: 
Try to factorize \(x^7 + x^2 + 1\).

Hint 2: 
Observe that  \(\omega\) and \(\omega^2 \) are the roots of \(x^7 + x^2 + 1\).

Hint 3:
\(x^7 + x^2 + 1\) =  (\(x^2 + x + 1\)).(\(x^5 - x^4 + x^2 -x + 1\))

A shorter solution or approach can always exist. Think about it. If you find an alternative solution or approach, mention it in the comments. We would love to hear something different from you.

5 comments on “2016 ISI Objective Solution Problem 1”

  1. I did it using the theory of equation.
    Sum of roots of the given polynomial x^7 + x^2 + 1 = 0
    if Option one is true then it must be multiplied to x^2 + x + 1 as the sum of roots of the polynomial in option one is 1. Multiplying x^5 - x^4 + x^2 -x + 1 with x^2+x+1 yields us the required polynomial.
    I was lucky that the first option itself was correct. But even if it wouid not have been then finding the pair and multilying would not have been that long as the given polynomials are easy to multiply

  2. By 1+w+w^2=0 , where 'w' is comlex cube root of unity. we see that 'w' and 'w^2' satisfy x^7+x^2+1=0. So x^2+x+1 is a factor.

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