To find the additional linear factors of the polynomial \( x^3 - 6x^2 + 11x - 6 \) given that \( x - 3 \) is a factor, we can perform polynomial long division or synthetic division of the cubic polynomial by \( x - 3 \).
Let's use synthetic division:
- Write down the coefficients of the polynomial: \( 1, -6, 11, -6 \).
- Use \( 3 \) (the root associated with the factor \( x - 3 \)) for synthetic division.
\[ \begin{array}{r|rrrr} 3 & 1 & -6 & 11 & -6 \ & & 3 & -9 & 6 \ \hline & 1 & -3 & 2 & 0 \ \end{array} \]
The row at the bottom gives us the coefficients of the quotient polynomial: \( 1, -3, 2 \), which corresponds to:
\[ x^2 - 3x + 2 \]
Now, we can factor \( x^2 - 3x + 2 \):
\[ x^2 - 3x + 2 = (x - 1)(x - 2) \]
Putting it all together, the original polynomial can be expressed as:
\[ x^3 - 6x^2 + 11x - 6 = (x - 3)(x - 1)(x - 2) \]
Thus, the additional linear factors are \( x - 1 \) and \( x - 2 \).
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