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.
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.
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
X = --2; the only problem is demonstrating uniqueness but that should be a breeze.
쉽긴 뭐가 쉽니? ㅋㅋㅋㅋ
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.
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 ^^
french curve keeps on saying pierce me
Please proceed faster! Do you think we’re idiots? 😋
-2
Браво-брависимо
x = - 2.
Браво-брависимо
@@АндрейКожевников-о8й
Thank you, it was so easy to see. There are three more solutions, though. ☺