Question
$$8x^4+4x^2-5x+\frac{1}{2}x = 0$$
Answer
$$ x = 0 $$
Explanation
$$ \begin{aligned} 8x^4+4x^2-5x+\frac{1}{2}x &= 0&& \text{multiply ALL terms by } \color{blue}{ 2 }. \\[1 em]2\cdot8x^4+2\cdot4x^2-2\cdot5x+2 \cdot \frac{1}{2}x &= 2\cdot0&& \text{cancel out the denominators} \\[1 em]16x^4+8x^2-10x+x &= 0&& \text{simplify left side} \\[1 em]16x^4+8x^2-9x &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 16x^{4}+8x^{2}-9x = 0 } $, first we need to factor our $ x $.
$$ 16x^{4}+8x^{2}-9x = x \left( 16x^{3}+8x-9 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 16x^{3}+8x-9 = 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