Solve x^3-8x^2+25x-26=0

$$x ^ { 3 } - 8 x ^ { 2 } + 25 x - 26 = 0$$

Solve for x (complex solution)

$$x=3-2i$$

$$x=2$$

$$x=3+2i$$

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 -26 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±26,±13,±2,±1$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-8x^{2}+25x-26 by x-2 to get x^{2}-6x+13. Solve the equation where the result equals to 0.

$$x^{2}-6x+13=0$$

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, -6 for b, and 13 for c in the quadratic formula.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 1\times 13}}{2}$$

Do the calculations.

$$x=\frac{6±\sqrt{-16}}{2}$$

Solve the equation x^{2}-6x+13=0 when ± is plus and when ± is minus.

$$x=3-2i x=3+2i$$

List all found solutions.

$$x=2 x=3-2i x=3+2i$$

Solve for x

$$x=2$$

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 -26 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±26,±13,±2,±1$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-8x^{2}+25x-26 by x-2 to get x^{2}-6x+13. Solve the equation where the result equals to 0.

$$x^{2}-6x+13=0$$

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, -6 for b, and 13 for c in the quadratic formula.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 1\times 13}}{2}$$

Do the calculations.

$$x=\frac{6±\sqrt{-16}}{2}$$

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.

$$x=2$$