$$\frac{6x^4+4x^3+2x^2+x-2}{3x^2+x-1} = 0$$
Answer
$$ \begin{matrix}x_1 = 0.51354 & x_2 = -0.92386 & x_3 = -0.12817+0.82835i \\[1 em] x_4 = -0.12817-0.82835i & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{6x^4+4x^3+2x^2+x-2}{3x^2+x-1} &= 0&& \text{multiply ALL terms by } \color{blue}{ 3x^2+x-1 }. \\[1 em](3x^2+x-1)\frac{6x^4+4x^3+2x^2+x-2}{3x^2+x-1} &= (3x^2+x-1)\cdot0&& \text{cancel out the denominators} \\[1 em]6x^4+4x^3+2x^2+x-2 &= 0&& \\[1 em] \end{aligned} $$
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