Solve 3x^4-14x^3-6x^2+6x-5=0

$$3 { x }^{ 4 } -14 { x }^{ 3 } -6 { x }^{ 2 } +6x-5=0$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{2}i+1}{3}\approx 0.333333333-0.471404521i$$

$$x=5$$

$$x=\frac{1+\sqrt{2}i}{3}\approx 0.333333333+0.471404521i$$

$$x=-1$$

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 -5 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.

$$±\frac{5}{3},±5,±\frac{1}{3},±1$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 3x^{4}-14x^{3}-6x^{2}+6x-5 by x+1 to get 3x^{3}-17x^{2}+11x-5. Solve the equation where the result equals to 0.

$$3x^{3}-17x^{2}+11x-5=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -5 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.

$$±\frac{5}{3},±5,±\frac{1}{3},±1$$

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.

$$x=5$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 3x^{3}-17x^{2}+11x-5 by x-5 to get 3x^{2}-2x+1. Solve the equation where the result equals to 0.

$$3x^{2}-2x+1=0$$

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 3 for a, -2 for b, and 1 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{2±\sqrt{-8}}{6}$$

Solve the equation 3x^{2}-2x+1=0 when ± is plus and when ± is minus.

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

List all found solutions.

$$x=-1 x=5 x=\frac{-\sqrt{2}i+1}{3} x=\frac{1+\sqrt{2}i}{3}$$

Solve for x

$$x=-1$$

$$x=5$$

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 -5 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.

$$±\frac{5}{3},±5,±\frac{1}{3},±1$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 3x^{4}-14x^{3}-6x^{2}+6x-5 by x+1 to get 3x^{3}-17x^{2}+11x-5. Solve the equation where the result equals to 0.

$$3x^{3}-17x^{2}+11x-5=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -5 and q divides the leading coefficient 3. List all candidates \frac{p}{q}.

$$±\frac{5}{3},±5,±\frac{1}{3},±1$$

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.

$$x=5$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 3x^{3}-17x^{2}+11x-5 by x-5 to get 3x^{2}-2x+1. Solve the equation where the result equals to 0.

$$3x^{2}-2x+1=0$$

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 3 for a, -2 for b, and 1 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{2±\sqrt{-8}}{6}$$

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.

$$x=-1 x=5$$