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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

## By Srijit Mukherjee

I Learn. I Dream. I Enjoy. I Share.

## 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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