$$\frac{1+x}{2}+\frac{3-x}{4} = (x+2)\cdot2(x+1)\cdot3(x+3)$$
Answer
$$ \begin{matrix}x_1 = -0.92397 & x_2 = -2.12173 & x_3 = -2.9543 \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{1+x}{2}+\frac{3-x}{4} &= (x+2)\cdot2(x+1)\cdot3(x+3)&& \text{multiply ALL terms by } \color{blue}{ 4 }. \\[1 em]4 \cdot \frac{1+x}{2}+4\frac{3-x}{4} &= 4(x+2)\cdot2(x+1)\cdot3(x+3)&& \text{cancel out the denominators} \\[1 em]2x+2+3-x &= 24x^3+144x^2+264x+144&& \text{simplify left side} \\[1 em]x+5 &= 24x^3+144x^2+264x+144&& \text{move all terms to the left hand side } \\[1 em]x+5-24x^3-144x^2-264x-144 &= 0&& \text{simplify left side} \\[1 em]-24x^3-144x^2-263x-139 &= 0&& \\[1 em] \end{aligned} $$
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