Question

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

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 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:

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