Question

What are the additional linear factors of x3−6x2+11x−6 if x−3 is a factor? (1 point)
Responses

(x+2)(x+1)
left parenthesis x plus 2 right parenthesis left parenthesis x plus 1 right parenthesis

(x−2)(x−1)
left parenthesis x minus 2 right parenthesis left parenthesis x minus 1 right parenthesis

(x−2)(x+1)
left parenthesis x minus 2 right parenthesis left parenthesis x plus 1 right parenthesis

(x+2)(x−1)

Answer 1

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 to divide the polynomial by \( x - 3 \).

1. Set up the long division with \( x^3 - 6x^2 + 11x - 6 \) divided by \( x - 3 \).

2. Divide the leading term:
- \( x^3 \div x = x^2 \)
- Multiply \( x^2 \) by \( x - 3 \) to get \( x^3 - 3x^2 \).

3. Subtract this from the original polynomial:
\[
(x^3 - 6x^2 + 11x - 6) - (x^3 - 3x^2) = -3x^2 + 11x - 6
\]

4. Repeat the process:
- \( -3x^2 \div x = -3x \)
- Multiply \( -3x \) by \( x - 3 \) to get \( -3x^2 + 9x \).
- Subtract:
\[
(-3x^2 + 11x - 6) - (-3x^2 + 9x) = 2x - 6
\]

5. Repeat again:
- \( 2x \div x = 2 \)
- Multiply \( 2 \) by \( x - 3 \) to get \( 2x - 6 \).
- Subtract:
\[
(2x - 6) - (2x - 6) = 0
\]

The result of the division is \( x^2 - 3x + 2 \).

Now factor \( x^2 - 3x + 2 \):
- This can be factored as \( (x - 1)(x - 2) \).

Thus, the complete factorization of \( x^3 - 6x^2 + 11x - 6 \) is:
\[
(x - 3)(x - 1)(x - 2)
\]

The additional linear factors of the polynomial are \( (x - 1)(x - 2) \), or in the response format:

**(x−2)(x−1)**.