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This question already has an answer here:

I am stuck at my exam practice here.

The remainder of the division of $x^3$ by $x^2-x+1$ is ..... and that of $x^{2007}$ by $x^2-x+1$ is .....

I tried the polynomial remainder theorem but I am not sure if I did it correctly.

By factor theorem definition, provided by Wikipedia,

the remainder of the division of a polynomial $f(x)$ by a linear polynomial $x-r$ is equal to $f(r)$.

So I attempted to find $r$ by factorizing $x^2-x+1$ first but I got the complex form $x=\frac{1\pm\sqrt{3}i}{2}=r$.

$f(r)$ is then $(\frac{1+\sqrt{3}i}{2})^3$ or $(\frac{1-\sqrt{3}i}{2})^3$ which do not sound right.

However, the answer key provided is $-1$ for the first question and also $-1$ for the second one. Please help.

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marked as duplicate by Bill Dubuque algebra-precalculus Jun 18 at 16:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ math.stackexchange.com/questions/3266201/… $\endgroup$ – lab bhattacharjee Jun 18 at 15:12
  • $\begingroup$ Why is my method using the remainder theorem $f(r)$ faulty though? $\endgroup$ – Trey Anupong Jun 18 at 15:18
  • $\begingroup$ Just to nitpick, what you provided there does not follow by definition, but rather, by the factor theorem. The definition of the remainder of $p/q$ is the polynomial $r$ such that $$\tfrac{p}{q}=s + \tfrac{r}{q},$$ where $s$ is polynomial and $\deg(r) < \deg(q)$. $\endgroup$ – Luke Collins Jun 18 at 15:21
  • $\begingroup$ @LukeCollins thank you for pointing that out, edited done! $\endgroup$ – Trey Anupong Jun 18 at 15:29
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    $\begingroup$ @TreyAnupong Then better to start with the 3rd link $\endgroup$ – Bill Dubuque Jun 18 at 16:39
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Since $$x^3+1 = (x+1)(x^2-x+1)$$ so $$x^3 = (x+1)(x^2-x+1)-1$$ the answer is $-1$.

Similarly for \begin{eqnarray}x^{3n}+1 &=& (x^3+1)\underbrace{\Big((x^3)^{n-1}-(x^3)^{n-2}+...-(x^3)+1\Big)}_{q(x)}\\ &=& (x+1)(x^2-x+1)q(x)\\ \end{eqnarray}

so the answer is again $-1$.

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Another way if you're familiar with modular arithmetic is to work modulo $x^2-x+1$, in which case we have $x^2\equiv x-1$ and thus $$x^3\equiv xx^2\equiv x(x-1)\equiv x^2-x\equiv (x-1)-x\equiv -1.$$ This can be extended to your other question by noting that $x^{2007}=\left(x^3\right)^{669}$.

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One way of writing this is to borrow the notion of equivalence (encoded in the notion of Ideals etc)

Because $x^3+1=(x+1)(x^2-x+1)$, we can write $x^3\equiv -1 \bmod (x^2-x+1)$

Then $x^{2007}=(x^3)^{669}\equiv (-1)^{669}$

This can be a surprisingly effective and efficient way of doing these questions about polynomial division and remainders.

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(1) Since $x^3=(x^3+1)-1=(x+1)(x^2-x+1)-1$, the remainder is $-1$.

(2) $\displaystyle x^{2007}=(x^3)^{669}=[(x+1)(x^2-x+1)-1]^{669}=-1+\sum_{k=1}^{669}(x+1)^k(x^2-x+1)^k(-1)^{669-k}$

The remainder is $-1$.

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