Question
$$7n^3-8n^2+3n^4-(6n^2+6n^3+2n) = 0$$
Answer
$$ n = 0 $$
Explanation
$$ \begin{aligned} 7n^3-8n^2+3n^4-(6n^2+6n^3+2n) &= 0&& \text{simplify left side} \\[1 em]7n^3-8n^2+3n^4-6n^2-6n^3-2n &= 0&& \\[1 em]3n^4+n^3-14n^2-2n &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 3x^{4}+x^{3}-14x^{2}-2x = 0 } $, first we need to factor our $ x $.
$$ 3x^{4}+x^{3}-14x^{2}-2x = x \left( 3x^{3}+x^{2}-14x-2 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 3x^{3}+x^{2}-14x-2 = 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