A cubic with three roots

#cubic#factoring#roots

Statement

Solve in $\mathbb{R}$ the equation
$x^3-6x^2+11x-6=0.$

Hint

Test small integers.

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Test. $x=1$ : $1-6+11-6=0$ . Root. $x=2$ : $8-24+22-6=0$ . Root. $x=3$ : $27-54+33-6=0$ . Root.

Factor. $x^3-6x^2+11x-6=(x-1)(x-2)(x-3)$ .

Conclusion. $x\in\{1,2,3\}$ .