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Prove-that-x-5-3x-4-17x-3-x-2-3x-17-cannot-be-factorized-completely-over-the-set-of-polynomials-with-integral-coefficients-




Question Number 110420 by Aina Samuel Temidayo last updated on 28/Aug/20
Prove that  x^5 −3x^4 −17x^3 −x^2 −3x+17 cannot be  factorized completely over the set of  polynomials with integral coefficients.
$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{x}^{\mathrm{5}} −\mathrm{3x}^{\mathrm{4}} −\mathrm{17x}^{\mathrm{3}} −\mathrm{x}^{\mathrm{2}} −\mathrm{3x}+\mathrm{17}\:\mathrm{cannot}\:\mathrm{be} \\ $$$$\mathrm{factorized}\:\mathrm{completely}\:\mathrm{over}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of} \\ $$$$\mathrm{polynomials}\:\mathrm{with}\:\mathrm{integral}\:\mathrm{coefficients}. \\ $$
Answered by 1549442205PVT last updated on 29/Aug/20
For this,it is enough to prove that  Polynomial P(x)= x^5 −3x^4 −17x^3 −x^2 −3x+17  has no any integer roots   Hence,it is enough to  prove that all divisors of 17 aren′t  roots of P(x).This is always true by  checking directly
$$\mathrm{For}\:\mathrm{this},\mathrm{it}\:\mathrm{is}\:\mathrm{enough}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{Polynomial}\:\mathrm{P}\left(\mathrm{x}\right)=\:\mathrm{x}^{\mathrm{5}} −\mathrm{3x}^{\mathrm{4}} −\mathrm{17x}^{\mathrm{3}} −\mathrm{x}^{\mathrm{2}} −\mathrm{3x}+\mathrm{17} \\ $$$$\mathrm{has}\:\mathrm{no}\:\mathrm{any}\:\mathrm{integer}\:\mathrm{roots}\: \\ $$$$\mathrm{Hence},\mathrm{it}\:\mathrm{is}\:\mathrm{enough}\:\mathrm{to} \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{all}\:\mathrm{divisors}\:\mathrm{of}\:\mathrm{17}\:\mathrm{aren}'\mathrm{t} \\ $$$$\mathrm{roots}\:\mathrm{of}\:\mathrm{P}\left(\mathrm{x}\right).\mathrm{This}\:\mathrm{is}\:\mathrm{always}\:\mathrm{true}\:\mathrm{by} \\ $$$$\mathrm{checking}\:\mathrm{directly} \\ $$
Commented by Aina Samuel Temidayo last updated on 29/Aug/20
Thanks, but how do you know it has  no integer roots?
$$\mathrm{Thanks},\:\mathrm{but}\:\mathrm{how}\:\mathrm{do}\:\mathrm{you}\:\mathrm{know}\:\mathrm{it}\:\mathrm{has} \\ $$$$\mathrm{no}\:\mathrm{integer}\:\mathrm{roots}? \\ $$
Commented by 1549442205PVT last updated on 30/Aug/20
Because all divisors aren′t roots of  P(x)=0
$$\mathrm{Because}\:\mathrm{all}\:\mathrm{divisors}\:\mathrm{aren}'\mathrm{t}\:\mathrm{roots}\:\mathrm{of} \\ $$$$\mathrm{P}\left(\mathrm{x}\right)=\mathrm{0} \\ $$
Commented by Aina Samuel Temidayo last updated on 30/Aug/20
What divisors?
$$\mathrm{What}\:\mathrm{divisors}? \\ $$

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