Question

What are the additional linear factors of x3−6x2+11x−6 if x−3 is a factor? (1 point) Choose one CORRECT answer from the following:
A. (x−2)(x+1)
B. (x−2)(x−1)
C. (x+2)(x−1)
D. (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. **Divide \(x^3 - 6x^2 + 11x - 6\) by \(x - 3\)**:

- Divide \(x^3\) by \(x\) to get \(x^2\).
- Multiply \(x^2\) by \(x - 3\) to get \(x^3 - 3x^2\).
- Subtract:
\[
(x^3 - 6x^2 + 11x - 6) - (x^3 - 3x^2) = -3x^2 + 11x - 6
\]

- Now, divide \(-3x^2\) by \(x\) to get \(-3x\).
- Multiply \(-3x\) by \(x - 3\) to get \(-3x^2 + 9x\).
- Subtract:
\[
(-3x^2 + 11x - 6) - (-3x^2 + 9x) = 2x - 6
\]

- Now, divide \(2x\) by \(x\) to get \(2\).
- Multiply \(2\) by \(x - 3\) to get \(2x - 6\).
- Subtract:
\[
(2x - 6) - (2x - 6) = 0
\]

The result of the division gives us:
\[
x^3 - 6x^2 + 11x - 6 = (x - 3)(x^2 - 3x + 2)
\]

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

Thus, we have:
\[
x^3 - 6x^2 + 11x - 6 = (x - 3)(x - 1)(x - 2)
\]

The additional linear factors therefore are \((x - 1)(x - 2)\).

So the correct choice is:
**B. \((x - 2)(x - 1)\)**.