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Question Number 147364 by henderson last updated on 20/Jul/21

hi, everybody !  T = x^5 +3x^2 +2 is reductible in Q ?

$$\boldsymbol{\mathrm{hi}},\:\boldsymbol{\mathrm{everybody}}\:! \\ $$$$\boldsymbol{\mathrm{T}}\:=\:\boldsymbol{{x}}^{\mathrm{5}} +\mathrm{3}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{2}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{reductible}}\:\boldsymbol{\mathrm{in}}\:\mathbb{Q}\:? \\ $$

Answered by Olaf_Thorendsen last updated on 20/Jul/21

T∈Q[X], T(x) = x^5 +3x^2 +2  T(x) = x^5 +3x^2 +2  It′s easy to verify with the sign of  T′(x) that T has only one real root.  Then T is not reductible in Q, in R.  Only in C.

$$\mathrm{T}\in\mathbb{Q}\left[\mathrm{X}\right],\:\mathrm{T}\left({x}\right)\:=\:{x}^{\mathrm{5}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2} \\ $$$$\mathrm{T}\left({x}\right)\:=\:{x}^{\mathrm{5}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2} \\ $$$$\mathrm{It}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{verify}\:\mathrm{with}\:\mathrm{the}\:\mathrm{sign}\:\mathrm{of} \\ $$$$\mathrm{T}'\left({x}\right)\:\mathrm{that}\:\mathrm{T}\:\mathrm{has}\:\mathrm{only}\:\mathrm{one}\:\mathrm{real}\:\mathrm{root}. \\ $$$$\mathrm{Then}\:\mathrm{T}\:\mathrm{is}\:\mathrm{not}\:\mathrm{reductible}\:\mathrm{in}\:\mathbb{Q},\:\mathrm{in}\:\mathbb{R}. \\ $$$$\mathrm{Only}\:\mathrm{in}\:\mathbb{C}. \\ $$

Answered by puissant last updated on 20/Jul/21

P  is  a  polynomial..  P(X)=a_n X^n +a_(n−1) X^(n−1) +....+a_1 X+a_0   ∃ p ( prime number)/    • p split a_0  , a_1  , ..... , a_(n−1)     • p no split a_n     • p^2  no split a_0   ⇒ P  is irreductible in Q[x]  (Einsentein theorem)...

$${P}\:\:{is}\:\:{a}\:\:{polynomial}.. \\ $$$${P}\left({X}\right)={a}_{{n}} {X}^{{n}} +{a}_{{n}−\mathrm{1}} {X}^{{n}−\mathrm{1}} +....+{a}_{\mathrm{1}} {X}+{a}_{\mathrm{0}} \\ $$$$\exists\:{p}\:\left(\:{prime}\:{number}\right)/ \\ $$$$\:\:\bullet\:{p}\:{split}\:{a}_{\mathrm{0}} \:,\:{a}_{\mathrm{1}} \:,\:.....\:,\:{a}_{{n}−\mathrm{1}} \\ $$$$\:\:\bullet\:{p}\:{no}\:{split}\:{a}_{{n}} \\ $$$$\:\:\bullet\:{p}^{\mathrm{2}} \:{no}\:{split}\:{a}_{\mathrm{0}} \\ $$$$\Rightarrow\:{P}\:\:{is}\:{irreductible}\:{in}\:{Q}\left[{x}\right] \\ $$$$\left({Einsentein}\:{theorem}\right)... \\ $$

Commented by Olaf_Thorendsen last updated on 20/Jul/21

The Eisenstein irreductibility  criterion is not verified here mister.

$${The}\:{Eisenstein}\:{irreductibility} \\ $$$${criterion}\:{is}\:{not}\:{verified}\:{here}\:{mister}. \\ $$

Commented by puissant last updated on 20/Jul/21

monsieur ca veut dire que le polynome  est reductible..

$${monsieur}\:{ca}\:{veut}\:{dire}\:{que}\:{le}\:{polynome} \\ $$$${est}\:{reductible}.. \\ $$

Commented by Olaf_Thorendsen last updated on 20/Jul/21

Non. Si le critere est verifie alors le  polynome est irreductible dans Q et  donc dans Z selon le lemme de Gauss.  Si le critere n′est pas respecte, on ne  peut rien deduire.  Par exemple 4x^2 −1 est reductible  dans Q mais ne respecte pas le  critere.

$${Non}.\:{Si}\:{le}\:{critere}\:{est}\:{verifie}\:{alors}\:{le} \\ $$$${polynome}\:{est}\:{irreductible}\:{dans}\:{Q}\:{et} \\ $$$${donc}\:{dans}\:{Z}\:{selon}\:{le}\:{lemme}\:{de}\:{Gauss}. \\ $$$${Si}\:{le}\:{critere}\:{n}'{est}\:{pas}\:{respecte},\:{on}\:{ne} \\ $$$${peut}\:{rien}\:{deduire}. \\ $$$${Par}\:{exemple}\:\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}\:{est}\:{reductible} \\ $$$${dans}\:{Q}\:{mais}\:{ne}\:{respecte}\:{pas}\:{le} \\ $$$${critere}. \\ $$

Commented by puissant last updated on 20/Jul/21

Okey monsieur je comprend maintenant  merci prof..

$${Okey}\:{monsieur}\:{je}\:{comprend}\:{maintenant} \\ $$$${merci}\:{prof}.. \\ $$

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