Solve x^3-7x^2+17x-15=0

$$x ^ { 3 } - 7 x ^ { 2 } + 17 x - 15 = 0$$

Solve for x (complex solution)

$$x=2-i$$

$$x=3$$

$$x=2+i$$

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 -15 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±15,±5,±3,±1$$

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-7x^{2}+17x-15 by x-3 to get x^{2}-4x+5. Solve the equation where the result equals to 0.

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

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 1 for a, -4 for b, and 5 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{4±\sqrt{-4}}{2}$$

Solve the equation x^{2}-4x+5=0 when ± is plus and when ± is minus.

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

List all found solutions.

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

Solve for x

$$x=3$$

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 -15 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±15,±5,±3,±1$$

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-7x^{2}+17x-15 by x-3 to get x^{2}-4x+5. Solve the equation where the result equals to 0.

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

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 1 for a, -4 for b, and 5 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{4±\sqrt{-4}}{2}$$

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.

$$x=3$$