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Post Closed as "Not suitable for this site" by Emil Jeřábek, Max Alekseyev, abx, Daniele Tampieri, Steven Landsburg
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toni_iva
toni_iva
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Question

I am interested in the root of the polynomial function :

$x^4+(a+b+c+d-2)x^3+(ab+ac+ad+bc+bd+cd-2b-2c-a-d)x^2 +(abc+abd+acd+bcd-ab-ac-ad-2bc-bd-cd-a-d+b+c)z+ abcd-abc-bcd-ad+bc=0$$x^4+(a+b+c+d-2)x^3+(ab+ac+ad+bc+bd+cd-2b-2c-a-d)x^2 +(abc+abd+acd+bcd-ab-ac-ad-2bc-bd-cd-a-d+b+c)x+ abcd-abc-bcd-ad+bc=0$

Under the restriction that: $0<a,b,c,d<1$.

I tried Vietas formula but don’t know how to come to a decomposition

$a_0=abcd-abc-bcd-ad+bc=x_1x_2x_3x_4$

Question

I am interested in the root of the polynomial function :

$x^4+(a+b+c+d-2)x^3+(ab+ac+ad+bc+bd+cd-2b-2c-a-d)x^2 +(abc+abd+acd+bcd-ab-ac-ad-2bc-bd-cd-a-d+b+c)z+ abcd-abc-bcd-ad+bc=0$

Under the restriction that: $0<a,b,c,d<1$.

I tried Vietas formula but don’t know how to come to a decomposition

$a_0=abcd-abc-bcd-ad+bc=x_1x_2x_3x_4$

Question

I am interested in the root of the polynomial function :

$x^4+(a+b+c+d-2)x^3+(ab+ac+ad+bc+bd+cd-2b-2c-a-d)x^2 +(abc+abd+acd+bcd-ab-ac-ad-2bc-bd-cd-a-d+b+c)x+ abcd-abc-bcd-ad+bc=0$

Under the restriction that: $0<a,b,c,d<1$.

I tried Vietas formula but don’t know how to come to a decomposition

$a_0=abcd-abc-bcd-ad+bc=x_1x_2x_3x_4$

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toni_iva
toni_iva
  • 39
  • 5

roots of a fourth degree polynomial function (Vieta)

Question

I am interested in the root of the polynomial function :

$x^4+(a+b+c+d-2)x^3+(ab+ac+ad+bc+bd+cd-2b-2c-a-d)x^2 +(abc+abd+acd+bcd-ab-ac-ad-2bc-bd-cd-a-d+b+c)z+ abcd-abc-bcd-ad+bc=0$

Under the restriction that: $0<a,b,c,d<1$.

I tried Vietas formula but don’t know how to come to a decomposition

$a_0=abcd-abc-bcd-ad+bc=x_1x_2x_3x_4$