Question

It is necessary to solve the equation!
HELP GUYS!

Answer 1

3x⁵ - 2x³ + 18x² - 12 = 0 ⇔ x³(3x² - 2) + 6(3x² - 2) = 0 ⇔

⇔(3x² - 2)(x³ + 6) = 0

3x² - 2 = 0 ⇒ x = ±√(2/3)

[tex] x^3 +6 = 0 \Rightarrow x^3+(\sqrt[3]6)^3 =0 \Rightarrow (x+\sqrt[3]6 )(x^2-x\sqrt[3]6+\sqrt[3]{6^2})=0 \\\;\\ x+\sqrt[3]6 =0 \Rightarrow x = -\sqrt[3]6 \\\;\\ x^2-x\sqrt[3]6+\sqrt[3]{6^2} =0 \Rightarrow x\in \mathbb{C}\backslash \mathbb{R}[/tex]

Answer 2

Radacinile:  x1,x2,x3 ∈R iar x4,x5 ∈ C.