What are the additional linear factors of x3−6x^2+11x−6 if x−3 is a factor?

1 answer

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:

  1. Write down the coefficients of the polynomial: \( 1, -6, 11, -6 \).
  2. 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:

\[ (x - 3)(x - 1)(x - 2) \]