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# 2016 ISI Objective Solution Problem 1

## 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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# STOP

### 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. ## By Srijit Mukherjee

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## 5 replies on “2016 ISI Objective Solution Problem 1” DEB JYOTI MITRAsays:

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 Srijit Mukherjeesays:

Yes, you are right!:) DEB MITRAsays:

thanks Ramesh Knowledge Indexsays:

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 Bangalore Mathematical Institutesays:

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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