Solve x^4-5x^3+8x^2-5x+1=0

$$x ^ { 4 } - 5 x ^ { 3 } + 8 x ^ { 2 } - 5 x + 1 = 0$$

Solve for x

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

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

$$x=1$$

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}-5x^{3}+8x^{2}-5x+1 by x-1 to get x^{3}-4x^{2}+4x-1. Solve the equation where the result equals to 0.

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

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}-4x^{2}+4x-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.

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

Do the calculations.

$$x=\frac{3±\sqrt{5}}{2}$$

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

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

List all found solutions.

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