What is the remainder when $P(x^{12})$ is divided by $P(x)$?

Question

Let $P(x) = 1 + x + x^2 + x^3+ x^4 + x^5$. What is the remainder when $P(x^{12})$ is divided by $P(x)$?
(a) $0$
(b) $6$
(c) $1+x$
(d) $1 + x + x^2 + x^3+ x^4 $

The way I adopted to solve this question was by putting x=1 which gives remainder= $0$ and this matches with option (a). But the answer to this question is 6. I am not able to understand the solution given:

It would be great if someone could explain the solution in simpler or more intuitive terms. Another way of solving the question is also welcome.

Inserting $x=1$ is not something you can do that easily. For instance, the division $\frac{x^2+5}{x}$ has a remainder of $5$, but inserting $x=1$ makes it $\frac 61$, which has no remainder. — Arthur, Oct 27 '16 at 08:05
This was wrongly closed as a dupe, which it is not since it is asking about an explanation of a specific proof (not given in the claimed dupe). Nor does the purported dupe explain how to correctly perform the method attempted. — Bill Dubuque, Oct 27 '16 at 17:10

score 4 · Accepted Answer · answered Oct 27 '16 at 08:04

4

$P(x) = \frac{x^6-1}{x-1}$ $$x^{6n}-1 = (x^6-1)(1+x^6+x^{12}+\cdots +x^{6(n-1)})$$ Thus $$P(x^{12}) - 6 = x^{60} -1 + x^{48} - 1 + x^{36}-1 + x^{24}-1 + x^{12}- 1 $$ and hence $P(x^{12})-6$ is a multiple of $x^6-1$ and hence $P(x)$ divides $P(x^{12})-6$. Thus the reminder is 6.

answered Oct 27 '16 at 08:04

This is much clearer using modular arithmetic - see my answer. – Bill Dubuque Oct 27 '16 at 17:12
@BillDubuque Yes, you are right. I wasn't sure whether OP will know modular arithmetic as applied in polynomial rings. – Oct 28 '16 at 00:23
It is easy to check that the proofs of the Congruence Arithmetic Rules still work. In fact they hold in any (commutative) ring, since they use only ring laws. – Bill Dubuque Oct 28 '16 at 00:41

score 2 · Answer 2 · edited Apr 13 '17 at 12:21

2

mod $\,P(x)\,$ we have $\,\color{#c00}{x^6\equiv \bf 1}\,$ (not $\,x\equiv 1)\,$ by $\,\color{#c00}{x^6\!-1} = (x\!-\!1)P(x)\equiv 0$

so $\ P(x^{12}) = P((\color{#c00}{x^6})^2)\equiv P(\color{#c00}{\bf 1}^2)\equiv P(1)\equiv 6\ $ by the Polynomial Congruence Rule.

edited Apr 13 '17 at 12:21

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answered Oct 27 '16 at 17:09

Bill Dubuque

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Bernard · Answer 3 · 2016-10-27T18:13:06.363

Maybe this variant will be more intuitive:

Polynomial division leads to an equality $$P(x^{12})=P(x)Q(x)+R(x)$$ where the remainder $R(x)$ has degree $\le 4$ or $R(x)=0$

The roots of the divisor $P(x)$ are the non-real $5$-th roots of unity, and for any such root $\alpha$, we have $$P(\alpha^{12})=P(\alpha)Q(\alpha)+R(\alpha)=R(\alpha).$$ One calculates that each $R(\alpha)$ is equal to $6$.

Now a theorem on polynomials asserts that a polynomial of degree $d$ is uniquely determined by its values at $d+1$ distinct points. So, $R(x)$, which coincides with the constant polynomial $6$ at $5$ points and has degree at most$4$, is necessarily equal to this constant.

What is the remainder when $P(x^{12})$ is divided by $P(x)$?

3 Answers3