Solve x^4+x^3-4x^2+x+1=0

$${ x }^{ 4 } + { x }^{ 3 } -4 { x }^{ 2 } +x+1 = 0$$

Solve for x

$$x=\frac{\sqrt{5}-3}{2}\approx -0.381966011$$

$$x=\frac{-\sqrt{5}-3}{2}\approx -2.618033989$$

$$x=1$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 1 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}+x^{3}-4x^{2}+x+1 by x-1 to get x^{3}+2x^{2}-2x-1. Solve the equation where the result equals to 0.

$$x^{3}+2x^{2}-2x-1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+2x^{2}-2x-1 by x-1 to get x^{2}+3x+1. Solve the equation where the result equals to 0.

$$x^{2}+3x+1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 3 for b, and 1 for c in the quadratic formula.

$$x=\frac{-3±\sqrt{3^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$x=\frac{-3±\sqrt{5}}{2}$$

Solve the equation x^{2}+3x+1=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{5}-3}{2} x=\frac{\sqrt{5}-3}{2}$$

List all found solutions.

$$x=1 x=\frac{-\sqrt{5}-3}{2} x=\frac{\sqrt{5}-3}{2}$$