$$\frac{1+x}{2}+\frac{3-x}{4} = x^3-7x^2+13x-7$$
Answer
$$ \begin{matrix}x_1 = 4.63187 & x_2 = 1.18406+0.61574i & x_3 = 1.18406-0.61574i \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{1+x}{2}+\frac{3-x}{4} &= x^3-7x^2+13x-7&& \text{multiply ALL terms by } \color{blue}{ 4 }. \\[1 em]4 \cdot \frac{1+x}{2}+4\frac{3-x}{4} &= 4x^3-4\cdot7x^2+4\cdot13x-4\cdot7&& \text{cancel out the denominators} \\[1 em]2x+2+3-x &= 4x^3-28x^2+52x-28&& \text{simplify left side} \\[1 em]x+5 &= 4x^3-28x^2+52x-28&& \text{move all terms to the left hand side } \\[1 em]x+5-4x^3+28x^2-52x+28 &= 0&& \text{simplify left side} \\[1 em]-4x^3+28x^2-51x+33 &= 0&& \\[1 em] \end{aligned} $$
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