помогите решить х^3+2x^2-4x-8=0. Created by eldariksanov77. algebra-ru

помогите решить х^3+2x^2-4x-8=0...

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$x^3+2x^2-4x-8=0 \\ x^2(x+2)-4(x+2)=0 \\ (x^2-4)(x+2)=0 \\ (x-2)(x+2)^2=0 \\ \\ 1) \\ x-2=0 \\ x=2 \\ \\ 2) \\ (x+2)^2=0 \\ x+2=0 \\ x=-2$

Ответ: -2; 2

1 votes Thanks 0

Vladislav006

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$x^3+2x^2-4x-8=0$

$x^{2} (x+2)-4(x+2)=0$

$(x+2)(x^{2}-4) =0$

$1) (x+2) =0 \\ x_1=-2$

$2) (x^{2}-4) =0 \\ x^{2} =4 \\ x_{2,3} = \pm2$

Корень х= - 2 повторяется, тогда

Ответе:
$x_{1,2} = \pm2$

2 votes Thanks 1

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