$$\frac{3x^5+4x^4-2x^3}{x^3+2x+1}+8x^2-4x+6 = 0$$
Answer
$$ \begin{matrix}x_1 = -0.46627 & x_2 = 0.39399+0.78417i & x_3 = 0.39399-0.78417i \\[1 em] x_4 = -0.16086+1.22191i & x_5 = -0.16086-1.22191i \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{3x^5+4x^4-2x^3}{x^3+2x+1}+8x^2-4x+6 &= 0&& \text{multiply ALL terms by } \color{blue}{ x^3+2x+1 }. \\[1 em](x^3+2x+1)\frac{3x^5+4x^4-2x^3}{x^3+2x+1}+(x^3+2x+1)\cdot8x^2-(x^3+2x+1)\cdot4x+(x^3+2x+1)\cdot6 &= (x^3+2x+1)\cdot0&& \text{cancel out the denominators} \\[1 em]3x^5+4x^4-2x^3+8x^5+16x^3+8x^2-(4x^4+8x^2+4x)+6x^3+12x+6 &= 0&& \text{simplify left side} \\[1 em]11x^5+4x^4+14x^3+8x^2-(4x^4+8x^2+4x)+6x^3+12x+6 &= 0&& \\[1 em]11x^5+4x^4+14x^3+8x^2-4x^4-8x^2-4x+6x^3+12x+6 &= 0&& \\[1 em]11x^5+14x^3-4x+6x^3+12x+6 &= 0&& \\[1 em]11x^5+20x^3+8x+6 &= 0&& \\[1 em] \end{aligned} $$
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