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How to split the polynomial .


How to find all elements of a factor group?Split groups and quasi-split groups.Definition of split of exact sequenceIrreducibility of a cubic polynomialDetermine the irreducibility of polynomial.How to find the minimal polynomial.polynomial with a root but no linear factorReducing polynomial modulo $n$Finding minimal polynomial when the straight equation must be reduced.Splitting field of polynomial is F8






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3












$begingroup$


How do I split $x^2-5 $ in $mathbbZ/5mathbbZ$ ? Since $0$ is a root I have $x $ as linear factor . How can I find the other linear factor ?










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$endgroup$







  • 1




    $begingroup$
    $x^2=xtimes x$
    $endgroup$
    – J. W. Tanner
    9 hours ago

















3












$begingroup$


How do I split $x^2-5 $ in $mathbbZ/5mathbbZ$ ? Since $0$ is a root I have $x $ as linear factor . How can I find the other linear factor ?










share|cite|improve this question









$endgroup$







  • 1




    $begingroup$
    $x^2=xtimes x$
    $endgroup$
    – J. W. Tanner
    9 hours ago













3












3








3





$begingroup$


How do I split $x^2-5 $ in $mathbbZ/5mathbbZ$ ? Since $0$ is a root I have $x $ as linear factor . How can I find the other linear factor ?










share|cite|improve this question









$endgroup$




How do I split $x^2-5 $ in $mathbbZ/5mathbbZ$ ? Since $0$ is a root I have $x $ as linear factor . How can I find the other linear factor ?







abstract-algebra number-theory






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asked 9 hours ago









AnabolicHorseAnabolicHorse

1708 bronze badges




1708 bronze badges







  • 1




    $begingroup$
    $x^2=xtimes x$
    $endgroup$
    – J. W. Tanner
    9 hours ago












  • 1




    $begingroup$
    $x^2=xtimes x$
    $endgroup$
    – J. W. Tanner
    9 hours ago







1




1




$begingroup$
$x^2=xtimes x$
$endgroup$
– J. W. Tanner
9 hours ago




$begingroup$
$x^2=xtimes x$
$endgroup$
– J. W. Tanner
9 hours ago










2 Answers
2






active

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8












$begingroup$

Note that $x^2-5 = x^2$, which is already factorised.






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$endgroup$




















    1












    $begingroup$

    Hint: Vieta's formulas still work. The sum of the two roots equals the (negative of the) coefficient of the first degree term.






    share|cite|improve this answer









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      2 Answers
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      2 Answers
      2






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      active

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      8












      $begingroup$

      Note that $x^2-5 = x^2$, which is already factorised.






      share|cite|improve this answer









      $endgroup$

















        8












        $begingroup$

        Note that $x^2-5 = x^2$, which is already factorised.






        share|cite|improve this answer









        $endgroup$















          8












          8








          8





          $begingroup$

          Note that $x^2-5 = x^2$, which is already factorised.






          share|cite|improve this answer









          $endgroup$



          Note that $x^2-5 = x^2$, which is already factorised.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 9 hours ago









          Patrick StevensPatrick Stevens

          29.4k5 gold badges29 silver badges75 bronze badges




          29.4k5 gold badges29 silver badges75 bronze badges























              1












              $begingroup$

              Hint: Vieta's formulas still work. The sum of the two roots equals the (negative of the) coefficient of the first degree term.






              share|cite|improve this answer









              $endgroup$

















                1












                $begingroup$

                Hint: Vieta's formulas still work. The sum of the two roots equals the (negative of the) coefficient of the first degree term.






                share|cite|improve this answer









                $endgroup$















                  1












                  1








                  1





                  $begingroup$

                  Hint: Vieta's formulas still work. The sum of the two roots equals the (negative of the) coefficient of the first degree term.






                  share|cite|improve this answer









                  $endgroup$



                  Hint: Vieta's formulas still work. The sum of the two roots equals the (negative of the) coefficient of the first degree term.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 9 hours ago









                  ArthurArthur

                  132k9 gold badges127 silver badges219 bronze badges




                  132k9 gold badges127 silver badges219 bronze badges



























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