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I am told that the remainder on division of a polynomial $p(z)$ by $z^3+z^2+z+1$ is $z^2-z+1$. I am also given that $p(1)=2$, and then asked to determine the remainder when $p(z)$ is divided by $z^4-1$.

I have expressed $p(z)$ as $p(z) = (z^3+z^2+z+1)(q(z)) + z^2-z+1$, for some polynomial $q$. And I found that $p(1)=2$ implies $q(1)=1/4$.

I then expressed $p(z)$ also as $p(z)=(z^4-1)(h(z)) + r(z)$, for some polynomial $h(z)$, where $r(z)$ is the remainder from the division of $p(z)$ by $z^4-1$.

I then divided $z^4-1$ by $z^3+z^2+z+1$ and found that $z^4-1 = (z-1)(z^3+z^2+z+1)$, so that I could express $p(z)$ as $p(z)= (z-1)(z^3+z^2+z+1)(h(z)) + r(z)$.

From then on I couldn't see how to proceed. Any hints on how to better approach this problem would be appreciated!

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We write the Euclidean division of $q(z)$ by $z-1$

$$q(z)=(z-1)u(z)+v$$ and since $q(1)=\frac14$ so $v=\frac14$ hence

\begin{align}p(z) &= (z^3+z^2+z+1)q(z) + z^2-z+1 \\&= (z^3+z^2+z+1)((z-1)u(z)+\frac14) + z^2-z+1\\&=(z^4-1)u(z)+\underbrace{z^3+\frac54z^2-\frac34z+\frac54}_{\text{the remainder}}\end{align}

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  • $\begingroup$ Now that is fantastic. Thank you Sami, I appreciate! $\endgroup$ – 2good4this Oct 28 '14 at 23:09
  • $\begingroup$ You're welcome. $\endgroup$ – user63181 Oct 28 '14 at 23:11
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Hint: Write $q(z)=(z-1)s(z)+c$

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  • $\begingroup$ Could you explain a bit more how I could use your hint? $\endgroup$ – 2good4this Oct 28 '14 at 22:52
  • $\begingroup$ Thank you Hagen, this is the same idea that Sami used in his response. Thanks! $\endgroup$ – 2good4this Oct 28 '14 at 23:10

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