Question Number 147364 by henderson last updated on 20/Jul/21
$$\boldsymbol{\mathrm{hi}},\:\boldsymbol{\mathrm{everybody}}\:! \\ $$$$\boldsymbol{\mathrm{T}}\:=\:\boldsymbol{{x}}^{\mathrm{5}} +\mathrm{3}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{2}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{reductible}}\:\boldsymbol{\mathrm{in}}\:\mathbb{Q}\:? \\ $$
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.](https://www.tinkutara.com/question/Q147386.png)
$$\mathrm{T}\in\mathbb{Q}\left[\mathrm{X}\right],\:\mathrm{T}\left({x}\right)\:=\:{x}^{\mathrm{5}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2} \\ $$$$\mathrm{T}\left({x}\right)\:=\:{x}^{\mathrm{5}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2} \\ $$$$\mathrm{It}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{verify}\:\mathrm{with}\:\mathrm{the}\:\mathrm{sign}\:\mathrm{of} \\ $$$$\mathrm{T}'\left({x}\right)\:\mathrm{that}\:\mathrm{T}\:\mathrm{has}\:\mathrm{only}\:\mathrm{one}\:\mathrm{real}\:\mathrm{root}. \\ $$$$\mathrm{Then}\:\mathrm{T}\:\mathrm{is}\:\mathrm{not}\:\mathrm{reductible}\:\mathrm{in}\:\mathbb{Q},\:\mathrm{in}\:\mathbb{R}. \\ $$$$\mathrm{Only}\:\mathrm{in}\:\mathbb{C}. \\ $$
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)...](https://www.tinkutara.com/question/Q147389.png)
$${P}\:\:{is}\:\:{a}\:\:{polynomial}.. \\ $$$${P}\left({X}\right)={a}_{{n}} {X}^{{n}} +{a}_{{n}−\mathrm{1}} {X}^{{n}−\mathrm{1}} +….+{a}_{\mathrm{1}} {X}+{a}_{\mathrm{0}} \\ $$$$\exists\:{p}\:\left(\:{prime}\:{number}\right)/ \\ $$$$\:\:\bullet\:{p}\:{split}\:{a}_{\mathrm{0}} \:,\:{a}_{\mathrm{1}} \:,\:…..\:,\:{a}_{{n}−\mathrm{1}} \\ $$$$\:\:\bullet\:{p}\:{no}\:{split}\:{a}_{{n}} \\ $$$$\:\:\bullet\:{p}^{\mathrm{2}} \:{no}\:{split}\:{a}_{\mathrm{0}} \\ $$$$\Rightarrow\:{P}\:\:{is}\:{irreductible}\:{in}\:{Q}\left[{x}\right] \\ $$$$\left({Einsentein}\:{theorem}\right)… \\ $$
Commented by Olaf_Thorendsen last updated on 20/Jul/21
$${The}\:{Eisenstein}\:{irreductibility} \\ $$$${criterion}\:{is}\:{not}\:{verified}\:{here}\:{mister}. \\ $$
Commented by puissant last updated on 20/Jul/21
$${monsieur}\:{ca}\:{veut}\:{dire}\:{que}\:{le}\:{polynome} \\ $$$${est}\:{reductible}.. \\ $$
Commented by Olaf_Thorendsen last updated on 20/Jul/21
$${Non}.\:{Si}\:{le}\:{critere}\:{est}\:{verifie}\:{alors}\:{le} \\ $$$${polynome}\:{est}\:{irreductible}\:{dans}\:{Q}\:{et} \\ $$$${donc}\:{dans}\:{Z}\:{selon}\:{le}\:{lemme}\:{de}\:{Gauss}. \\ $$$${Si}\:{le}\:{critere}\:{n}'{est}\:{pas}\:{respecte},\:{on}\:{ne} \\ $$$${peut}\:{rien}\:{deduire}. \\ $$$${Par}\:{exemple}\:\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}\:{est}\:{reductible} \\ $$$${dans}\:{Q}\:{mais}\:{ne}\:{respecte}\:{pas}\:{le} \\ $$$${critere}. \\ $$
Commented by puissant last updated on 20/Jul/21
$${Okey}\:{monsieur}\:{je}\:{comprend}\:{maintenant} \\ $$$${merci}\:{prof}.. \\ $$