Solve x^3+x^2+x=14

$${ x }^{ 3 } + { x }^{ 2 } +x=14$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{19}i-3}{2}\approx -1.5-2.179449472i$$

$$x=2$$

$$x=\frac{-3+\sqrt{19}i}{2}\approx -1.5+2.179449472i$$

Solution Steps

Subtract 14 from both sides.

$$x^{3}+x^{2}+x-14=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -14 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±14,±7,±2,±1$$

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}+x^{2}+x-14 by x-2 to get x^{2}+3x+7. Solve the equation where the result equals to 0.

$$x^{2}+3x+7=0$$

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, 3 for b, and 7 for c in the quadratic formula.

$$x=\frac{-3±\sqrt{3^{2}-4\times 1\times 7}}{2}$$

Do the calculations.

$$x=\frac{-3±\sqrt{-19}}{2}$$

Solve the equation x^{2}+3x+7=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{19}i-3}{2} x=\frac{-3+\sqrt{19}i}{2}$$

List all found solutions.

$$x=2 x=\frac{-\sqrt{19}i-3}{2} x=\frac{-3+\sqrt{19}i}{2}$$

Solve for x

$$x=2$$

Solution Steps

Subtract 14 from both sides.

$$x^{3}+x^{2}+x-14=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -14 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±14,±7,±2,±1$$

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}+x^{2}+x-14 by x-2 to get x^{2}+3x+7. Solve the equation where the result equals to 0.

$$x^{2}+3x+7=0$$

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, 3 for b, and 7 for c in the quadratic formula.

$$x=\frac{-3±\sqrt{3^{2}-4\times 1\times 7}}{2}$$

Do the calculations.

$$x=\frac{-3±\sqrt{-19}}{2}$$

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.

$$x=2$$