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 to divide the polynomial by \( x - 3 \).
Let's perform synthetic division with the root \( x = 3 \):
3 | 1 -6 11 -6
| 3 -9 6
--------------------
1 -3 2 0
The result of the synthetic division is \( x^2 - 3x + 2 \), and the remainder is 0, confirming that \( x - 3 \) is indeed a factor. Thus, we can express the polynomial as:
\[ x^3 - 6x^2 + 11x - 6 = (x - 3)(x^2 - 3x + 2) \]
Next, we need to factor \( x^2 - 3x + 2 \):
To factor \( x^2 - 3x + 2 \), we look for two numbers that multiply to \( 2 \) (the constant term) and add up to \(-3\) (the coefficient of \( x \)). The numbers \(-2\) and \(-1\) work:
\[ x^2 - 3x + 2 = (x - 2)(x - 1) \]
Therefore, the complete factorization of the original polynomial can be expressed as:
\[ x^3 - 6x^2 + 11x - 6 = (x - 3)(x - 2)(x - 1) \]
The additional linear factors of the polynomial, besides \( x - 3 \), are:
\[ (x - 2)(x - 1) \]
Thus, the answer is:
\[ \text{(x−2)(x−1)} \]