Question Number 179900 by Shrinava last updated on 03/Nov/22
![Prove that: f = 5x^4 + 35x^3 + 28x^2 + 14x + 7 is irreducible in Q[x]](Q179900.png)
$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{f}\:=\:\mathrm{5x}^{\mathrm{4}} \:+\:\mathrm{35x}^{\mathrm{3}} \:+\:\mathrm{28x}^{\mathrm{2}} \:+\:\mathrm{14x}\:+\:\mathrm{7}\:\:\mathrm{is}\:\mathrm{irreducible}\:\mathrm{in}\:\:\mathrm{Q}\left[\mathrm{x}\right] \\ $$
Answered by floor(10²Eta[1]) last updated on 04/Nov/22
![we want a prime p such that 1) p∣35, p∣28, p∣14, p∣7 2) p∤5 3) p^2 ∤7 ∴p=7 works⇒f(x) is irreductible in Q[x]](Q179904.png)
$$\mathrm{we}\:\mathrm{want}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{p}\:\mathrm{such}\:\mathrm{that} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{p}\mid\mathrm{35},\:\mathrm{p}\mid\mathrm{28},\:\mathrm{p}\mid\mathrm{14},\:\mathrm{p}\mid\mathrm{7} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{p}\nmid\mathrm{5} \\ $$$$\left.\mathrm{3}\right)\:\mathrm{p}^{\mathrm{2}} \nmid\mathrm{7} \\ $$$$\therefore\mathrm{p}=\mathrm{7}\:\mathrm{works}\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{irreductible}\:\mathrm{in}\:\mathrm{Q}\left[\mathrm{x}\right] \\ $$