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The quartic factor is the most difficult. We get rid of this one with the substitution y=x+3. We turn it into the simpler y^4+y^3-2y^2-y+1=0. By Theorem, we know that the only possible candidates for integer roots are +1 and -1, which in fact both solve. Undoing the substitution we obtain that x=-4 or x=-2 (this is the one that several youtubers have already detected by simple inspection). From here, already knowing this, by simple polynomial division we convert it into y^2+y-1 to obtain the set of the 2 missing conjugate irrationals.
Remplazando x+1 por alguna variable, por ejemplo x+1=a. Luego efectuas las potencias y te saldra a^4 +8a^3+26a^2+34a+15=0 de ahi lo factorizas con ruffini y obtienes (a+1)(a^3+7a^2+19a+15) donde a+1=0, a=-1, reviertes el cambio de variable y x+1=-1, x=-2.
Remplazando x+1 por alguna variable, por ejemplo x+1=a. Luego efectuas las potencias y te saldra a^4 +8a^3+26a^2+34a+15=0 de ahi lo factorizas con ruffini y obtienes (a+1)(a^3+7a^2+19a+15) donde a+1=0, a=-1, reviertes el cambio de variable y x+1=-1, x=-2.
