Solve x^3-6x^2+15x-10=0

$$x ^ { 3 } - 6 x ^ { 2 } + 15 x - 10 = 0$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{15}i+5}{2}\approx 2.5-1.936491673i$$

$$x=1$$

$$x=\frac{5+\sqrt{15}i}{2}\approx 2.5+1.936491673i$$

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 -10 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±10,±5,±2,±1$$

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 x^{3}-6x^{2}+15x-10 by x-1 to get x^{2}-5x+10. Solve the equation where the result equals to 0.

$$x^{2}-5x+10=0$$

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, -5 for b, and 10 for c in the quadratic formula.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\times 10}}{2}$$

Do the calculations.

$$x=\frac{5±\sqrt{-15}}{2}$$

Solve the equation x^{2}-5x+10=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{15}i+5}{2} x=\frac{5+\sqrt{15}i}{2}$$

List all found solutions.

$$x=1 x=\frac{-\sqrt{15}i+5}{2} x=\frac{5+\sqrt{15}i}{2}$$

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 -10 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±10,±5,±2,±1$$

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 x^{3}-6x^{2}+15x-10 by x-1 to get x^{2}-5x+10. Solve the equation where the result equals to 0.

$$x^{2}-5x+10=0$$

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, -5 for b, and 10 for c in the quadratic formula.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\times 10}}{2}$$

Do the calculations.

$$x=\frac{5±\sqrt{-15}}{2}$$

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.

$$x=1$$