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(A) $ x^5 - x^4 + x^2 -x +1 \quad$ (B) $ x^5 + x^4 +1 \quad$ (C) $ x^5 + x^4 + x^2 +x +1\quad$

(D) $ x^5 - x^4 + x^2 +x +1$

My effort: Looking at the polynomial, I know that it will have only one real root, which is negative. All other 6 roots should be imaginary. And that's how it is in the four options as well. Without actually dividing each polynomials in the options, which looks a tedious way, is there a method for finding it?

Monalisha Bhowmik
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  • It's not that tedious! $x^7+x^2+1=(x^5-x^4+x^2-x+1)(x^2+\cdots)$. Only two more terms to find out. – Chrystomath Sep 16 '20 at 08:17
  • In general, no, there is no shortcut to actually dividing the two polynomials. But it's not *that* tedious, is it? – 5xum Sep 16 '20 at 08:17
  • Does this answer your question : [see here](https://math.stackexchange.com/a/2012353/811225) , and [this](https://math.stackexchange.com/questions/2012344/prime-factor-of-a-1471421/2012353#comment4131367_2012353) – Anindya Prithvi Sep 16 '20 at 08:22
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    @AnindyaPrithvi That thread was already linked in my answer. – Bill Dubuque Sep 16 '20 at 08:25
  • @BillDubuque This is not a "flag"...I saw your answer right now....should I remove this? I have linked a comment and an answer which is in direct relation – Anindya Prithvi Sep 16 '20 at 08:27
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    @AnindyaPrithvi No, my comment was merely meant to be informative (e.g. so readers don't waste time chasing the link twice). – Bill Dubuque Sep 16 '20 at 08:34

4 Answers4

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Clearly $\omega,{\omega}^2$ satisfy.Therefore $x^2+x+1$ divides It is easy to find the other 5th degree polynomial(long division)

$\omega$ is cube root of unity

Albus Dumbledore
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Hint $\ \ {\begin{align} \ \ \{\color{#c00}{2,\ \ 1,\ \ 0}\}&\equiv\, \{\color{#0a0}{A,\ B,\ C}\}\pmod{\!3}\\[.4em] \Longrightarrow\ \ x^{\large\color{#c00}2}\!\!+\!x^{\large\color{#c00}1}\!\!+\!x^{\large\color{#c00}0}&\mid\, x^{\color{#0a0}A}\! +\! x^{\color{#0a0}B}\! +\! x^\color{#0a0}{C}\end{align}}$

Bill Dubuque
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$$x^7+x^2+1=x^7-x+x^2+x+1=$$ $$=(x^2+x+1)(x(x-1)(x^3+1)+1)=$$ $$=(x^2+x+1)(x^5-x^3+x^2-x+1).$$

Michael Rozenberg
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Polynomial $x^7+x^2+1$ has higher term coefficient $1$ and also constant term coefficient $1$.

Thus if we divide it by a polynomial of degree $5$ of the form $(x^5+\cdots+1)$ as $A,B,C,D$ proposition are, it will be $x^2+ax+1$, only the central coefficient is unknown, the other two are forced to $1$.

If you multiply this polynomial by $B$ or $C$ $\begin{cases}(x^2+ax+1)(x^5+x^4+1)\\(x^2+ax+1)(x^5+x^4+x^3+x^2+1)\end{cases}$

you can notice that there will be an unremovable (and not desired) term in $x^4$, thus these possibilities should be excluded.

So you have to decide between $A$ and $D$ $\begin{cases}(x^2+ax+1)(x^5-x^4+x^2-x+1)\\(x^2+ax+1)(x^5-x^4+x^2+x+1)\end{cases}$

In both case we have $(a-1)x^6$ therefore $a=1$ to get rid of it.

The term in $x^1$ is also easy to get $A:(a-1)x=0$ and $D:(a+1)x=2x$, and since there should be no term in $x$, then $D$ should be excluded.

You just need to verify that $A$ works via a final expansion, which is indeed the case.

zwim
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