Solve x^3+2x^2+x=18

$$x ^ { 3 } + 2 x ^ { 2 } + x = 18$$

Solve for x (complex solution)

$$x=-\sqrt{5}i-2\approx -2-2.236067977i$$

$$x=2$$

$$x=-2+\sqrt{5}i\approx -2+2.236067977i$$

Solution Steps

Subtract 18 from both sides.

$$x^{3}+2x^{2}+x-18=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -18 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±18,±9,±6,±3,±2,±1$$

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}+2x^{2}+x-18 by x-2 to get x^{2}+4x+9. Solve the equation where the result equals to 0.

$$x^{2}+4x+9=0$$

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 9 for c in the quadratic formula.

$$x=\frac{-4±\sqrt{4^{2}-4\times 1\times 9}}{2}$$

Do the calculations.

$$x=\frac{-4±\sqrt{-20}}{2}$$

Solve the equation x^{2}+4x+9=0 when ± is plus and when ± is minus.

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

List all found solutions.

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

Solve for x

$$x=2$$

Solution Steps

Subtract 18 from both sides.

$$x^{3}+2x^{2}+x-18=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -18 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±18,±9,±6,±3,±2,±1$$

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}+2x^{2}+x-18 by x-2 to get x^{2}+4x+9. Solve the equation where the result equals to 0.

$$x^{2}+4x+9=0$$

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 9 for c in the quadratic formula.

$$x=\frac{-4±\sqrt{4^{2}-4\times 1\times 9}}{2}$$

Do the calculations.

$$x=\frac{-4±\sqrt{-20}}{2}$$

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.

$$x=2$$