Solve x³ − 6x² + 11x − 6 = 0.

Step-by-step solution

1. 1

   List the candidate rational roots. For a monic polynomial with integer coefficients, any rational root must divide the constant term $-6$:

   $\pm 1,\ \pm 2,\ \pm 3,\ \pm 6$

   The positive candidates are the likely ones here, since all sign changes in the coefficient string $1, -6, 11, -6$ suggest positive roots (Descartes rule of signs allows three or one).

2. 2

   Find the first root by substitution. Try $x = 1$:

   $1 - 6 + 11 - 6 = 0 \ \checkmark$

   So $x = 1$ is a root and, by the factor theorem, $(x - 1)$ divides the cubic exactly.

3. 3

   Divide out the known factor with synthetic division.

   $\begin{array}{r|rrrr} 1 & 1 & -6 & 11 & -6 \\ & & 1 & -5 & 6 \\ \hline & 1 & -5 & 6 & 0 \end{array}$

   The zero remainder confirms the root, and the bottom row gives the reduced quadratic:

   $x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6)$

4. 4

   Factor the quadratic. Two numbers multiplying to $6$ and adding to $-5$ are $-2$ and $-3$:

   $x^2 - 5x + 6 = (x - 2)(x - 3)$

   so the complete factorisation is

   $(x-1)(x-2)(x-3) = 0$

5. 5

   Apply the zero-product property and check. A product is zero only if a factor is zero, so

   $x = 1, \quad x = 2, \quad x = 3$

   Check each in the original: $8 - 24 + 22 - 6 = 0$ and $27 - 54 + 33 - 6 = 0$. As a further check, the roots must sum to $-(-6)/1 = 6$ (Vieta), and $1 + 2 + 3 = 6$; their product must be $-(-6)/1 = 6$, and $1 \cdot 2 \cdot 3 = 6$.