Dividing polynomial $f(x^3)$ by $x^2+x+1$

Question

Given two polynomials with real coefficients $g(x)$ and $h(x)$.

Additionally, $x^2+x+1$ is a factor of $f(x^3)= h(x^3)+xg(x^3)$.

Prove that $x-1|h(x)$ and $x-1|g(x)$

I have tried to solve it by doing this;

$(x-1)(x^2+x+1)|(x-1)(h(x^3)+xg(x^3))$

then; $(x^3-1)|(x)(h(x^3)-g(x^3)) + (x^2)g(x^3) - h(x^3)$

consider degree of $3k, 3k+1, 3k+2$ separately, I got;

$(x^3-1)|(x^2)g(x^3)$ and $h(x^3)$

Lastly, $(x^3-1)|g(x^3), h(x^3)$ and so on finish proving.

I wonder, is it correct to separate three cases of degrees in this problem?

For ploynomial $p(x)$ divisibility $(x-1)\mid p(x)$ means that $p(1)=0$. — richrow, May 04 '20 at 07:04
Related (USAMO1976#5): https://artofproblemsolving.com/community/c6h342993p1834510 — Batominovski, May 04 '20 at 18:41
@WETutorialSchool That too follows by one line of trivial mod arithmetic - same as in my answer. This and related problems all become trivial once one learns the general method of simpler multiples. — Bill Dubuque, May 04 '20 at 21:10
while the ideas are the same, I wouldn't say this is a duplicate of the cited question. — robjohn, May 05 '20 at 06:16

score 3 · Accepted Answer · answered May 04 '20 at 07:18

I'm not sure what you mean by that separate cases.

But here is how I did it.

Let $a$ and $b$ be roots for $x^2+x+1=0$, then $a^3=b^3=1$ and clearly $a\ne b$ since the discriminat is not $0$.

Since $$x^2+x+1\mid f(x)$$ we see that $f(a)= f(b)=0$

Now put

$$x=a:\;\;\; f(a^3) = h(a^3)+ag(a^3)$$

so $$ 0 = f(1) = h(1)+ag(1)$$ and similary for $b$:$$ 0 = h(1)+bg(1)$$

Sunbstract does eqautions: $$(a-b)g(1)=0\implies g(1)=0$$ and thus also $h(1)=0$. This means $x-1\mid g(x)$ and $x-1\mid h(x)$.

Thanks! - I haven't thought of using $a$ and $b$ as roots. – insidenothingisnothing May 04 '20 at 10:46 — insidenothingisnothing, May 04 '20 at 10:46

score 2 · Answer 2 · answered May 04 '20 at 17:50

2

$\!\bmod x^2\!+\!x\!+\!1\!:\,\ \color{#c00}{x^3\equiv 1}\,$ so $\, 0\equiv h(\color{#c00}{x^3})\!+\!xg(\color{#c00}{x^3})\equiv h(\color{#c00}1)+xg(\color{#c00}1)\iff h(1)=0=g(1)$

answered May 04 '20 at 17:50

Bill Dubuque

272,048

More generally see the method of simpler multiples – Bill Dubuque May 04 '20 at 17:51

robjohn · Answer 3 · 2020-05-04T13:34:20.550

Suppose $$ \left.x^2+x+1\,\,\middle|\,\,h\!\left(x^3\right)+x\,g\!\left(x^3\right)\right.\tag1 $$ Let ${\bar h}(x)=h(x+1)$ and ${\bar g}(x)=g(x+1)$, then $(1)$ becomes $$ \left.x^2+x+1\,\,\middle|\,\,{\bar h}\!\left(x^3-1\right)+x\,{\bar g}\!\left(x^3-1\right)\right.\tag2 $$ Write $\bar h$ and $\bar g$ as $$ \bar h(x)=\sum_{k=0}^n\bar h_kx^k\tag3 $$ and $$ \bar g(x)=\sum_{k=0}^n\bar g_kx^k\tag4 $$ Since $\left.x^2+x+1\,\middle|\,x^3-1\right.$, we immediately have that $x^2+x+1$ divides each term in the sum for ${\bar h}\!\left(x^3-1\right)$ except for $\bar h_0$. Similarly, $x^2+x+1$ divides each term in the sum for ${\bar g}\!\left(x^3-1\right)$ except for $\bar g_0$. Therefore, $(2)$ implies $$ \left.x^2+x+1\,\,\middle|\,\,{\bar h}_0+x\,{\bar g}_0\right.\tag5 $$ which means that ${\bar h}_0={\bar g}_0=0$; that is, the constant coefficients of $\bar h$ and $\bar g$ are $0$. Thus, $$ \left.x-1\,\middle|\,\bar h(x-1)=h(x)\right.\tag6 $$ and $$ \left.x-1\,\middle|\,\bar g(x-1)=g(x)\right.\tag7 $$

Dividing polynomial $f(x^3)$ by $x^2+x+1$

3 Answers3