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

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

Solve for x (complex solution)

$$x=-1-i$$

$$x=\frac{1}{3}\approx 0.333333333$$

$$x=-1+i$$

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 -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.

$$x=\frac{1}{3}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 3x^{3}+5x^{2}+4x-2 by 3\left(x-\frac{1}{3}\right)=3x-1 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.

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

Do the calculations.

$$x=\frac{-2±\sqrt{-4}}{2}$$

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

$$x=-1-i x=-1+i$$

List all found solutions.

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

Solve for x

$$x=\frac{1}{3}\approx 0.333333333$$

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 -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.

$$x=\frac{1}{3}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 3x^{3}+5x^{2}+4x-2 by 3\left(x-\frac{1}{3}\right)=3x-1 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.

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

Do the calculations.

$$x=\frac{-2±\sqrt{-4}}{2}$$

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.

$$x=\frac{1}{3}$$