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Given the polynomials $f,g,h \in \mathbb{R}\left[X\right]$ with $$f=(x-1)^n-x^n+1$$ $$g=x^2-3x+2$$ $$h=x^2-x$$ where $n\ge3$

Find the remainder of dividing the polynomial f to g. Prove that if $n$ is odd, $h$ divides $f$ with no remainder.

I noticed that the three polynomials have a common root in $x=1$ meaning I can write all three of them like $(x+1)*p(x)$, thus having no remainder. I don't know if this helps me though. How do I solve this?

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  • $\begingroup$ For the second, you need to show that $x$ and $x-1$ divide $f$. To do this it is enough to show that $f(0)=0$ and $f(1)=0$. $\endgroup$ – André Nicolas Jul 17 '15 at 19:10
  • $\begingroup$ Thanks for the reply! Very insightful approach and solution, would've never thought of it. Any clue on the first one though? $\endgroup$ – MikhaelM Jul 17 '15 at 19:17
  • $\begingroup$ A simple way. Let the remainder by $ax+b$. So $f(x)=q(x)(x-1)(x-2)+ax+b$. Plug in $1$. We have $f(1)=0$ so $a+b=0$. Plug in $2$. We have $f(2)=2-2^n$ so $a+2b=2-2^n$. Solve for $a$ and $b$, easy. $\endgroup$ – André Nicolas Jul 17 '15 at 20:49
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Use the remainder theorem: for any polynomial $p(x)$, the remainder when dividing it by $x - a$ is given by $p(a)$.

Notice that $f(1) = 0$, so $x - 1$ divides evenly into $f(x)$. Therefore, for some $p(x)$, we may write, $$f(x) = (x - 1)p(x).$$ Further, since $f(2) = p(2)$ is the common remainder when dividing both $p(x)$ and $f(x)$ by $2$, we can write, $$p(x) = q(x)(x - 2) + p(2) = q(x)(x - 2) + f(2),$$ for some $q(x)$. Therefore, $$f(x) = (x - 1)(q(x)(x - 2) + f(2)) = (x^2 - 3x + 2)q(x) + f(2)(x - 1).$$ When dividing $f(x)$ by $x^2 - 3x + 2$, we therefore get a remainder of, $$f(2)(x - 1) = (2 - 2^n)(x - 1).$$

The second question is far more simple. Since $x^2 - x = x(x - 1)$, we just need to individually verify that $x$ and $x - 1$ divide evenly into $f(x)$ when $n$ is even. We do this by verifying $f(0) = f(1) = 0$. Notice that $f(1) = 0$ still, regardless of the value of $n$, but, $$f(0) = (0 - 1)^n - 0^n + 1 = (-1)^n + 1 = 0,$$ precisely when $n$ is odd, and we are done.

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    $\begingroup$ Thank you! Shouldn't last sentence be when $n$ is odd thought? Since $(-1)^n=-1$ only when $n$ is odd. $\endgroup$ – MikhaelM Jul 17 '15 at 19:33
  • $\begingroup$ Ah yes, thanks! $\endgroup$ – Theo Bendit Jul 18 '15 at 6:15
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For the first one:

Let $Q(x)$ and $R(x)$ be the quotient and remainder respectively when dividing $f(x)$ by $g(x)$.

Then, we have $f(x) = Q(x)g(x)+R(x)$ for all $x \in \mathbb{R}$.

Since $g(1) = 0$, plugging in $x = 1$ gives us $f(1) = R(1)$, i.e. $R(1) = 0$.

Since $g(2) = 0$, plugging in $x = 2$ gives us $f(2) = R(2)$, i.e. $R(2) = -2^n+2$.

Also, since $g$ has degree $2$, $R$ has degree at most $1$. Can you figure out what $R(x)$ is from this?


For the second one, the same method will work, except you should substitute the values of $x$ which are roots of $h(x)$ instead of $g(x)$.

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