Question

What are the additional linear factors of x3−6x2+11x−6
𝑥
3
−
6
𝑥
2
+
11
𝑥
−
6
if x−3
𝑥
−
3
is a factor? (1 point)
Responses

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

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

(x+2)(x+1)
(
𝑥
+
2
)
(
𝑥
+
1
)
left parenthesis x plus 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 use polynomial division or synthetic division to divide the polynomial by \( x - 3 \).

1. **Synthetic Division**: We perform synthetic division of \( x^3 - 6x^2 + 11x - 6 \) by \( x - 3 \).

```
3 | 1 -6 11 -6
| 3 -9 6
-------------------
1 -3 2 0
```

The result is \( x^2 - 3x + 2 \), with a remainder of 0, confirming that \( x - 3 \) is indeed a factor.

2. **Factoring the Result**: Now, we need to factor \( x^2 - 3x + 2 \).

The quadratic \( x^2 - 3x + 2 \) can be factored by finding two numbers that multiply to \( 2 \) (the constant term) and add to \( -3 \) (the coefficient of \( x \)). Those numbers are \( -1 \) and \( -2 \).

Thus, we can factor:

\[
x^2 - 3x + 2 = (x - 1)(x - 2)
\]

3. **Combining the Factors**: Therefore, the full factorization of \( x^3 - 6x^2 + 11x - 6 \) is:

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

From the responses given, the additional linear factors (aside from \( x - 3 \)) are:

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

Thus, the correct answer is:

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