X*x*x-x*x-8x+12=

To solve the polynomial equation \( x^3 - x^2 - 8x + 12 = 0 \), we can start by looking for rational roots using the Rational Root Theorem. The possible rational roots could be the factors of the constant term (12) divided by the factors of the leading coefficient (1).

The factors of 12 are ±1, ±2, ±3, ±4, ±6, ±12.

Let's test these values to see if any of them are roots.

1. Testing \( x = 1 \): \[ 1^3 - 1^2 - 8 \cdot 1 + 12 = 1 - 1 - 8 + 12 = 4 \quad (\text{not a root}) \]

2. Testing \( x = 2 \): \[ 2^3 - 2^2 - 8 \cdot 2 + 12 = 8 - 4 - 16 + 12 = 0 \quad (\text{is a root}) \]

Since \( x = 2 \) is a root, we can factor the polynomial \( x^3 - x^2 - 8x + 12 \) using synthetic division or polynomial long division.

Dividing \( x^3 - x^2 - 8x + 12 \) by \( x - 2 \):

\[ \begin{array}{r|rrrr} 2 & 1 & -1 & -8 & 12 \\ & & 2 & 2 & -12 \\ \hline & 1 & 1 & -6 & 0 \\ \end{array} \]

The result is \( x^2 + x - 6 \). Now we can factor this quadratic:

\[ x^2 + x - 6 = (x - 2)(x + 3) \]

Thus, the original polynomial can be factored as:

\[ x^3 - x^2 - 8x + 12 = (x - 2)(x - 2)(x + 3) = (x - 2)^2(x + 3) \]

Setting this equal to zero gives the roots:

\[ (x - 2)^2 = 0 \quad \Rightarrow \quad x = 2 \quad (\text{with multiplicity 2}) \] \[ (x + 3) = 0 \quad \Rightarrow \quad x = -3 \]

In conclusion, the roots of the equation \( x^3 - x^2 - 8x + 12 = 0 \) are:

\[ x = 2 \quad (\text{multiplicity 2}), \quad x = -3 \]