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?

  • $\begingroup$ 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

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.

  • 1
    $\begingroup$ 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

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 $g(x)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.