The polynomial \(x^7+x^2+1\) is divisible by
The problem is easy to understand. Your first task is to try the problem yourself.
Try to factorize \(x^7 + x^2 + 1\).
Observe that \(\omega\) and \(\omega^2 \) are the roots of \(x^7 + x^2 + 1\).
\(x^7 + x^2 + 1\) = (\(x^2 + x + 1\)).(\(x^5 - x^4 + x^2 -x + 1\))
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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.