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

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

Solve for x (complex solution)

$$x=-\frac{\sqrt{35}i}{10}+\frac{1}{2}\approx 0.5-0.591607978i$$

$$x=-1$$

$$x=\frac{\sqrt{35}i}{10}+\frac{1}{2}\approx 0.5+0.591607978i$$

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

$$±\frac{3}{5},±3,±\frac{1}{5},±1$$

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

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

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

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

Do the calculations.

$$x=\frac{5±\sqrt{-35}}{10}$$

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

$$x=-\frac{\sqrt{35}i}{10}+\frac{1}{2} x=\frac{\sqrt{35}i}{10}+\frac{1}{2}$$

List all found solutions.

$$x=-1 x=-\frac{\sqrt{35}i}{10}+\frac{1}{2} x=\frac{\sqrt{35}i}{10}+\frac{1}{2}$$

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

$$±\frac{3}{5},±3,±\frac{1}{5},±1$$

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

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

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

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

Do the calculations.

$$x=\frac{5±\sqrt{-35}}{10}$$

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.

$$x=-1$$