Solve x^4-5x-6=0

$${ x }^{ 4 } -5x-6=0$$

Solve for x (complex solution)

$$x=-1$$

$$x=2$$

$$x=\frac{-\sqrt{11}i-1}{2}\approx -0.5-1.658312395i$$

$$x=\frac{-1+\sqrt{11}i}{2}\approx -0.5+1.658312395i$$

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 -6 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±6,±3,±2,±1$$

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}-5x-6 by x+1 to get x^{3}-x^{2}+x-6. Solve the equation where the result equals to 0.

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -6 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±6,±3,±2,±1$$

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

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

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 3 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{-1±\sqrt{-11}}{2}$$

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

$$x=\frac{-\sqrt{11}i-1}{2} x=\frac{-1+\sqrt{11}i}{2}$$

List all found solutions.

$$x=-1 x=2 x=\frac{-\sqrt{11}i-1}{2} x=\frac{-1+\sqrt{11}i}{2}$$

Solve for x

$$x=2$$

$$x=-1$$

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 -6 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±6,±3,±2,±1$$

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}-5x-6 by x+1 to get x^{3}-x^{2}+x-6. Solve the equation where the result equals to 0.

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -6 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±6,±3,±2,±1$$

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

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

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 3 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{-1±\sqrt{-11}}{2}$$

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.

$$x=-1 x=2$$