Solve 3x^4-11x^3-3x^2-6x+8=0

$$3 x ^ { 4 } - 11 x ^ { 3 } - 3 x ^ { 2 } - 6 x + 8 = 0$$

Solve for x (complex solution)

$$x=\frac{2}{3}\approx 0.666666667$$

$$x=4$$

$$x=\frac{-\sqrt{3}i-1}{2}\approx -0.5-0.866025404i$$

$$x=\frac{-1+\sqrt{3}i}{2}\approx -0.5+0.866025404i$$

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 8 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.

$$±\frac{8}{3},±8,±\frac{4}{3},±4,±\frac{2}{3},±2,±\frac{1}{3},±1$$

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.

$$x=4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 3x^{4}-11x^{3}-3x^{2}-6x+8 by x-4 to get 3x^{3}+x^{2}+x-2. Solve the equation where the result equals to 0.

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

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 3. List all candidates \frac{p}{q}.

$$±\frac{2}{3},±2,±\frac{1}{3},±1$$

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.

$$x=\frac{2}{3}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 3x^{3}+x^{2}+x-2 by 3\left(x-\frac{2}{3}\right)=3x-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.

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

Do the calculations.

$$x=\frac{-1±\sqrt{-3}}{2}$$

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

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

List all found solutions.

$$x=4 x=\frac{2}{3} x=\frac{-\sqrt{3}i-1}{2} x=\frac{-1+\sqrt{3}i}{2}$$

Solve for x

$$x=\frac{2}{3}\approx 0.666666667$$

$$x=4$$

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 8 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.

$$±\frac{8}{3},±8,±\frac{4}{3},±4,±\frac{2}{3},±2,±\frac{1}{3},±1$$

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.

$$x=4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 3x^{4}-11x^{3}-3x^{2}-6x+8 by x-4 to get 3x^{3}+x^{2}+x-2. Solve the equation where the result equals to 0.

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

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 3. List all candidates \frac{p}{q}.

$$±\frac{2}{3},±2,±\frac{1}{3},±1$$

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.

$$x=\frac{2}{3}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 3x^{3}+x^{2}+x-2 by 3\left(x-\frac{2}{3}\right)=3x-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.

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

Do the calculations.

$$x=\frac{-1±\sqrt{-3}}{2}$$

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.

$$x=4 x=\frac{2}{3}$$