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Find a monic quartic polynomial $f(x)$ with rational coefficients whose roots include $x=3-i\sqrt[4]2$. Give your answer in expanded form


Would I use some kind of factoring of $(x+y)^4$ I am guessing that there will a step to get rid of the imaginary value, and then a step to get rid of the irrational values, but I do not know how to approach this... Can someone please give me a hint?

Thanks!

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    $\begingroup$ what is $(x-3)^4 \; ? \;$ $\endgroup$
    – Will Jagy
    Feb 21 '20 at 19:58
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    $\begingroup$ Use conjugates; e.g., multiply $3-i\sqrt[4]2$ by $3+i\sqrt[4]2$; but Will Jagy’s hint is better $\endgroup$ Feb 21 '20 at 20:00
  • $\begingroup$ $(x-3)^4=x^4+12x^3+54x^2+108x+81$ How does this help us...? $\endgroup$
    – Mike Smith
    Feb 21 '20 at 23:23
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$$ \begin{aligned} x&= 3 - i \sqrt[4]{2} \\ x-3&= -i \sqrt[4]{2} \\ (x-3)^4&= 2 \\ (x-3)^4-2&=0 \\ x^4 - 12 x^3 + 54 x^2 - 108 x + 79&=0 \end{aligned} $$

EDIT. In fact, you can even factor the given polynomial fully (if you desired) because examining the construction closely you can see the roots of $x^4 - 12 x^3 + 54 x^2 - 108 x + 79$ are precisely, $3 \pm i \sqrt[4]{2}$ and $3 \pm \sqrt[4]{2}$, i.e. $3 + \sqrt[4]{2} e^{2\pi ki/4}$ for $k=0,1,2,3$.

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  • $\begingroup$ wow, thank you, this helps me understand how to do these a lot better. $\endgroup$
    – Mike Smith
    Feb 21 '20 at 23:25

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