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

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

Solve for x (complex solution)

$$x=-\frac{\sqrt{51}i}{6}-\frac{1}{2}\approx -0.5-1.190238071i$$

$$x=1$$

$$x=\frac{\sqrt{51}i}{6}-\frac{1}{2}\approx -0.5+1.190238071i$$

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

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

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, 3 for b, and 5 for c in the quadratic formula.

$$x=\frac{-3±\sqrt{3^{2}-4\times 3\times 5}}{2\times 3}$$

Do the calculations.

$$x=\frac{-3±\sqrt{-51}}{6}$$

Solve the equation 3x^{2}+3x+5=0 when ± is plus and when ± is minus.

$$x=-\frac{\sqrt{51}i}{6}-\frac{1}{2} x=\frac{\sqrt{51}i}{6}-\frac{1}{2}$$

List all found solutions.

$$x=1 x=-\frac{\sqrt{51}i}{6}-\frac{1}{2} x=\frac{\sqrt{51}i}{6}-\frac{1}{2}$$

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

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

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, 3 for b, and 5 for c in the quadratic formula.

$$x=\frac{-3±\sqrt{3^{2}-4\times 3\times 5}}{2\times 3}$$

Do the calculations.

$$x=\frac{-3±\sqrt{-51}}{6}$$

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.

$$x=1$$