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x^{3}-6x^{2}+11x-6=0
Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.
±6,±3,±2,±1
By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -6 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.
x=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}-5x+6=0
By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-6x^{2}+11x-6 by x-1 to get x^{2}-5x+6. Solve the equation where the result equals to 0.
x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\times 6}}{2}
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, -5 for b, and 6 for c in the quadratic formula.
x=\frac{5±1}{2}
Do the calculations.
x=2 x=3
Solve the equation x^{2}-5x+6=0 when ± is plus and when ± is minus.
x=1 x=2 x=3
List all found solutions.