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How can I know if $ \ \ 2(t-1) \ \ $ divides $ \ \ t^4+2t^3-2t^2-3t+2$ over $\mathbb Q$ ? I don't want to use the polynomial division algorithm.

My approach : $(t-1)$ divides the polynomial since $1$ is a zero of it. Now my doubt is if $2$ divides the polynomial. My idea was that asking this is equivalent to asking if $ 2| t^4-3t$ since the other coefficients of the polynomials are even.

sigmatau
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    Over $\mathbb{Q}$, of course $2$ divides any polynomial, since it's a unit. You have $t^4 - 3t = 2(\frac{1}{2}t^4 - \frac{3}{2}t)$. Over $\mathbb{Z}$, this isn't true, though. – Cocopuffs Jul 29 '13 at 18:02

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HINT:

$$t^4-3t=t^4-t^3+t^3-t^2+t^2-3t=t^3(t-1)+t^2(t-1)+t(t-1)+2t$$

Each term except the last is divisible by $t(t-1)$ which is even

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Obviously $2$ divides $t^4-3t$

because $t^4-3t=(t^4-t)-2t=t(t^3-1)-2t=t(t-1)(t^2+t+1)-2t$

As $t(t-1)$ divisible by $2$ it implies that $t^4-3t$ divisible by $2$

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Clearly $t-1$ divides the polynomial (As you've already found why). Now note that $$t^4+2t^3-2t^2-3t+2\equiv t(t^3-1)\mod 2\equiv t(t-1) \mod 2\equiv0 \mod 2$$