$$\frac{8x^3+5x^2+3x-1}{2x} = 0$$
Answer
$$ \begin{matrix}x_1 = 0 & x_2 = 0.22201 & x_3 = -0.4235+0.61942i \\[1 em] x_4 = -0.4235-0.61942i & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{8x^3+5x^2+3x-1}{2x} &= 0&& \text{multiply ALL terms by } \color{blue}{ 2x }. \\[1 em]2x \cdot \frac{8x^3+5x^2+3x-1}{2x} &= 2x\cdot0&& \text{cancel out the denominators} \\[1 em]8x^5+5x^4+3x^3-x^2 &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 8x^{5}+5x^{4}+3x^{3}-x^{2} = 0 } $, first we need to factor our $ x^2 $.
$$ 8x^{5}+5x^{4}+3x^{3}-x^{2} = x^2 \left( 8x^{3}+5x^{2}+3x-1 \right) $$$ x = 0 $ is a root of multiplicity $ 2 $.
The remaining roots can be found by solving equation $ 8x^{3}+5x^{2}+3x-1 = 0$.
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using qubic formulas.
This page was created using
Polynomial Equations Solver