Solve x^4+x^3-6x^2-2x+4=0

$$x ^ { 4 } + x ^ { 3 } - 6 x ^ { 2 } - 2 x + 4 = 0$$

Solve for x

$$x=-\sqrt{3}-1\approx -2.732050808$$

$$x=2$$

$$x=\sqrt{3}-1\approx 0.732050808$$

$$x=-1$$

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

$$±4,±2,±1$$

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}+x^{3}-6x^{2}-2x+4 by x+1 to get x^{3}-6x+4. Solve the equation where the result equals to 0.

$$x^{3}-6x+4=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 4 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±4,±2,±1$$

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}-6x+4 by x-2 to get x^{2}+2x-2. Solve the equation where the result equals to 0.

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

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, 2 for b, and -2 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{-2±2\sqrt{3}}{2}$$

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

$$x=-\sqrt{3}-1 x=\sqrt{3}-1$$

List all found solutions.

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