Question

WHAT ARE THE ADDITIONAL LINEAR FACTORS x^3-6x^2+11x-6

Answer 1

To factor the polynomial \( x^3 - 6x^2 + 11x - 6 \), we can start by looking for its roots using the Rational Root Theorem. This theorem suggests that any potential rational root is a factor of the constant term (\(-6\)) divided by a factor of the leading coefficient (\(1\)). The possible rational roots are \( \pm 1, \pm 2, \pm 3, \pm 6 \).

We'll test these values by substituting them into the polynomial:

1. **Testing \( x = 1 \)**:
\[
1^3 - 6(1^2) + 11(1) - 6 = 1 - 6 + 11 - 6 = 0
\]
Therefore, \( x = 1 \) is a root.

Next, we can use synthetic division to divide the polynomial by \( x - 1 \).

Using synthetic division with root \( 1 \):

```
1 | 1 -6 11 -6
| 1 -5 6
------------------
1 -5 6 0
```

The result is \( x^2 - 5x + 6 \) with a remainder of \( 0\).

Now, we can factor the resulting quadratic polynomial:

\[
x^2 - 5x + 6 = (x - 2)(x - 3)
\]

Thus, we can express the original polynomial as:

\[
x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3)
\]

The additional linear factors in the polynomial \( x^3 - 6x^2 + 11x - 6 \) are \( (x - 2) \) and \( (x - 3) \). Therefore, the complete factorization is:

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