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.

Thank you. But we can do a little differently ! (1) x=y-3 . (2) y^4+y^3-2*y^2-y+1=0 ; divide both sides of the equations by (y)^2=no=0 : (3) y^2+1/y^2+y-1/y-2=0 ; (4) y-1/y=t : (5) y^2+1/y^2=t^2+2 ; (6) t^2+t=0 ; (7) t=0 , (8) t=-1 ; (7) - (4) : y+1/y=0 - (9) y=+-1 ; (8)- (4) : y+1/y=-1 - (10) y=[-1+-sqrt(5) ]/2 ; (9) - (1) and (10) - (1) : we get your answer !! With respect , Lidiy

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.

X = --2; the only problem is demonstrating uniqueness but that should be a breeze.

@thebooldang-death

6 ай бұрын

쉽긴 뭐가 쉽니? ㅋㅋㅋㅋ

@arturciyo8340

29 күн бұрын

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.

x+3=t и всё будет гораздо проще.

Solving time can be reduced further.

Pascal' triangle is better to be start with 1 1 at the 1st degree ^^

x = - 2.

@user-lf1bm8fm2g

6 ай бұрын

Браво-брависимо

@toveirenestrand3547

6 ай бұрын

@@user-lf1bm8fm2g Thank you, it was so easy to see. There are three more solutions, though. ☺

-2

@user-lf1bm8fm2g

6 ай бұрын

Браво-брависимо