$$\frac{3x^5+4x^4-2x^3}{x^2+2x+1}+8x^2-4x+6 = 0$$
Answer
$$ \begin{matrix}x_1 = -3.0517 & x_2 = 0.35604+0.79862i & x_3 = 0.35604-0.79862i \\[1 em] x_4 = -0.8302+0.40984i & x_5 = -0.8302-0.40984i \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{3x^5+4x^4-2x^3}{x^2+2x+1}+8x^2-4x+6 &= 0&& \text{multiply ALL terms by } \color{blue}{ x^2+2x+1 }. \\[1 em](x^2+2x+1)\frac{3x^5+4x^4-2x^3}{x^2+2x+1}+(x^2+2x+1)\cdot8x^2-(x^2+2x+1)\cdot4x+(x^2+2x+1)\cdot6 &= (x^2+2x+1)\cdot0&& \text{cancel out the denominators} \\[1 em]3x^5+4x^4-2x^3+8x^4+16x^3+8x^2-(4x^3+8x^2+4x)+6x^2+12x+6 &= 0&& \text{simplify left side} \\[1 em]3x^5+12x^4+14x^3+8x^2-(4x^3+8x^2+4x)+6x^2+12x+6 &= 0&& \\[1 em]3x^5+12x^4+14x^3+8x^2-4x^3-8x^2-4x+6x^2+12x+6 &= 0&& \\[1 em]3x^5+12x^4+10x^3-4x+6x^2+12x+6 &= 0&& \\[1 em]3x^5+12x^4+10x^3+6x^2+8x+6 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using Newton method.
This page was created using
Polynomial Equations Solver