The polynomial $ x^7 + x^2 +1$ is divisible by

Question

(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?

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
@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
@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

Albus Dumbledore · Answer 1 · 2020-09-16T08:21:22.160

3

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

edited Sep 16 '20 at 08:21

answered Sep 16 '20 at 08:19

Albus Dumbledore

11,860

Bill Dubuque · Answer 2 · 2020-09-16T16:55:41.267

1

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

edited Sep 16 '20 at 16:55

answered Sep 16 '20 at 08:20

Bill Dubuque

272,048

Note $,x^{\large\color{#c00}0} = 1,$ and $\ P\mid Q,$ means $P$ divides $Q\ $ (standard number theory notation) – Bill Dubuque Sep 16 '20 at 16:56

Michael Rozenberg · Answer 3 · 2020-09-16T09:13:10.177

0

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

edited Sep 16 '20 at 09:13

answered Sep 16 '20 at 08:36

Michael Rozenberg

194,933

score 0 · Answer 4 · answered Sep 16 '20 at 09:04

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.

The polynomial $ x^7 + x^2 +1$ is divisible by

4 Answers4