Solve x^3-3x=2

$${ x }^{ 3 } -3x=2$$

Solve for x

$$x=-1$$

$$x=2$$

Solution Steps

Subtract 2 from both sides.

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -2 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±2,±1$$

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

$$x^{2}-x-2=0$$

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, -1 for b, and -2 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{1±3}{2}$$

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

$$x=-1 x=2$$

List all found solutions.

$$x=-1 x=2$$