Solve the cubic equation x^3 - 6x^2 + 11x - 6 = 0.

Step-by-step solution

1. 1

   List the candidate rational roots. The polynomial is monic with constant term $-6$, so by the rational root theorem any rational root is an integer dividing $6$:

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

   All coefficients alternate in sign with a positive leading term, so by Descartes' rule the negative candidates cannot work — test the positive ones first.

2. 2

   Test $x=1$.

   $1-6+11-6=0$

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

3. 3

   Divide out the known factor by synthetic division. With coefficients $1,\ -6,\ 11,\ -6$ and divisor $x-1$: bring down $1$; $1\cdot 1=1$, $-6+1=-5$; $-5\cdot 1=-5$, $11-5=6$; $6\cdot 1=6$, $-6+6=0$. The zero remainder confirms the root, and the quotient is

   $x^{2}-5x+6$

   so $x^{3}-6x^{2}+11x-6=(x-1)\left(x^{2}-5x+6\right)$.

4. 4

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

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

   The complete factorisation is $(x-1)(x-2)(x-3)$.

5. 5

   Apply the zero-product property. A product is zero exactly when a factor is:

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

6. 6

   Verify all three and sanity-check with Vieta. Substituting gives $0$ at each of $1$, $2$, $3$. Vieta also matches: the roots sum to $1+2+3=6$ (the negative of the $x^{2}$ coefficient), pair-sum to $2+3+6=11$, and multiply to $1\cdot 2\cdot 3=6$.