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

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

Solve for x

$$x=\frac{1-\sqrt{5}}{2}\approx -0.618033989$$

$$x=2$$

$$x = \frac{\sqrt{5} + 1}{2} \approx 1.618033989$$

$$x=-1$$

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 2 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±2,±1$$

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

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 2 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±2,±1$$

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.

$$x=2$$

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

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

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, -1 for b, and -1 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{1±\sqrt{5}}{2}$$

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

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

List all found solutions.

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