$$5b-3b^3-8b^2-(2b^3-4b^2+6b-3b^4) = 0$$
Answer
$$ \begin{matrix}b_1 = 0 & b_2 = 2.30719 & b_3 = -0.32026+0.20472i \\[1 em] b_4 = -0.32026-0.20472i & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 5b-3b^3-8b^2-(2b^3-4b^2+6b-3b^4) &= 0&& \text{simplify left side} \\[1 em]5b-3b^3-8b^2-2b^3+4b^2-6b+3b^4 &= 0&& \\[1 em]3b^4-5b^3-4b^2-b &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 3x^{4}-5x^{3}-4x^{2}-x = 0 } $, first we need to factor our $ x $.
$$ 3x^{4}-5x^{3}-4x^{2}-x = x \left( 3x^{3}-5x^{2}-4x-1 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 3x^{3}-5x^{2}-4x-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