$$3x^2(x-1)(x+1)-x^3+3 = 0$$
Answer
$$ \begin{matrix}x_1 = 0.95468+0.43505i & x_2 = 0.95468-0.43505i & x_3 = -0.78801+0.53625i \\[1 em] x_4 = -0.78801-0.53625i & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 3x^2(x-1)(x+1)-x^3+3 &= 0&& \text{simplify left side} \\[1 em](3x^3-3x^2)(x+1)-x^3+3 &= 0&& \\[1 em]3x^4+3x^3-3x^3-3x^2-x^3+3 &= 0&& \\[1 em]3x^4+3x^3-3x^3-3x^2-x^3+3 &= 0&& \\[1 em]3x^4-x^3-3x^2+3 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using quartic formulas
This page was created using
Polynomial Equations Solver