Solve 0=x^3-3x^2+2

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

Solve for x

$$x=\sqrt{3}+1\approx 2.732050808$$

$$x=1-\sqrt{3}\approx -0.732050808$$

$$x=1$$

Solution Steps

Swap sides so that all variable terms are on the left hand side.

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

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

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

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

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

Do the calculations.

$$x=\frac{2±2\sqrt{3}}{2}$$

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

$$x=1-\sqrt{3} x=\sqrt{3}+1$$

List all found solutions.

$$x=1 x=1-\sqrt{3} x=\sqrt{3}+1$$