To find the additional linear factors of the polynomial \( x^3 - 6x^2 + 11x - 6 \) given that \( x - 3 \) is a factor, we can use polynomial long division or synthetic division to divide the polynomial by \( x - 3 \).
Step 1: Perform Synthetic Division
Using synthetic division with 3:
- Write down the coefficients of the polynomial: \( 1, -6, 11, -6 \).
- Set up synthetic division:
3 | 1 -6 11 -6
| 3 -9 6
------------------------
1 -3 2 0
Step 2: Write the Result
The bottom row gives us the coefficients of the quotient polynomial \( x^2 - 3x + 2 \). Therefore, we have:
\[ x^3 - 6x^2 + 11x - 6 = (x - 3)(x^2 - 3x + 2) \]
Step 3: Factor \( x^2 - 3x + 2 \)
Next, we can factor \( x^2 - 3x + 2 \):
To factor \( x^2 - 3x + 2 \), we need two numbers that multiply to \( 2 \) (the constant term) and add up to \( -3 \) (the coefficient of \( -3x \)). These numbers are \( -1 \) and \( -2 \).
Thus, we can factor it as:
\[ x^2 - 3x + 2 = (x - 1)(x - 2) \]
Final Result
Substituting back, we have:
\[ x^3 - 6x^2 + 11x - 6 = (x - 3)(x - 1)(x - 2) \]
Conclusion
The additional linear factors of \( x^3 - 6x^2 + 11x - 6 \) are \( x - 1 \) and \( x - 2 \). Thus, the complete factorization is:
\[ (x - 3)(x - 1)(x - 2) \]
1 answer