Solve x^4-2x^3+6x^2+22x+13=0

$$x ^ { 4 } - 2 x ^ { 3 } + 6 x ^ { 2 } + 22 x + 13 = 0$$

Solve for x (complex solution)

$$x=2+3i$$

$$x=2-3i$$

$$x=-1$$

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

$$±13,±1$$

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}+6x^{2}+22x+13 by x+1 to get x^{3}-3x^{2}+9x+13. Solve the equation where the result equals to 0.

$$x^{3}-3x^{2}+9x+13=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 13 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±13,±1$$

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}-3x^{2}+9x+13 by x+1 to get x^{2}-4x+13. Solve the equation where the result equals to 0.

$$x^{2}-4x+13=0$$

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, -4 for b, and 13 for c in the quadratic formula.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\times 13}}{2}$$

Do the calculations.

$$x=\frac{4±\sqrt{-36}}{2}$$

Solve the equation x^{2}-4x+13=0 when ± is plus and when ± is minus.

$$x=2-3i x=2+3i$$

List all found solutions.

$$x=-1 x=2-3i x=2+3i$$

Solve for x

$$x=-1$$

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

$$±13,±1$$

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}+6x^{2}+22x+13 by x+1 to get x^{3}-3x^{2}+9x+13. Solve the equation where the result equals to 0.

$$x^{3}-3x^{2}+9x+13=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 13 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±13,±1$$

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}-3x^{2}+9x+13 by x+1 to get x^{2}-4x+13. Solve the equation where the result equals to 0.

$$x^{2}-4x+13=0$$

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, -4 for b, and 13 for c in the quadratic formula.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\times 13}}{2}$$

Do the calculations.

$$x=\frac{4±\sqrt{-36}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=-1$$