Problem
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:
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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:
\(x^7 + x^2 + 1\) = (\(x^2 + x + 1\)).(\(x^5 – x^4 + x^2 -x + 1\))
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Related
5 replies on “2016 ISI Objective Solution Problem 1”
f(x)=(x^7+x^2+1);
g(x)=(x^2+x+1);
we observe
f(x)/g(x)=(x^5-x^4+x^2-x+1) using WolframAlpha
Yes, you are right!:)
thanks
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
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.