Question

решите уравнение 2х³+х²-8х-4=0​

Answer 1

$2x^{3} + x^{2} - 8x - 4 = 0\\\\x^{2} * (2x+1) - 4(2x+1)=0\\\\(2x + 1) * (x^{2} - 4) = 0\\\\2x + 1 = 0\\x^{2} - 4 = 0\\\\x = - \frac{1}{2} \\\\x = -2\\x = 2\\\\Otvet:\\x1 = -2 , x2 = - \frac{1}{2} , x3 = 2$

Answer 2

Ответ:

-0,5; 2; -2

Объяснение:

$2x^3+x^2-8x-4=0\\x^2(2x+1)-4(2x+1)=0\\(2x+1)(x^2-4)=0\\(2x+1)(x-2)(x+2)=0\\2x+1=0\;\; \; V \; \; x-2=0\; \;V \;\;x+2=0\\x_1=-0,5\; \; \;\;\;\; \; \;x_2=2\; \; \;\; \; \; \; \; \;\; \; \;x_3=-2$