Question

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

Answer 1

To find the additional linear factors of the polynomial \( x^3 - 6x^2 + 11x - 6 \), we can start by factoring the polynomial completely.

First, we can use the Rational Root Theorem to test for possible rational roots, which are the factors of the constant term (-6) over the factors of the leading coefficient (1). The possible rational roots are \( \pm 1, \pm 2, \pm 3, \pm 6 \).

Let's test these values:

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

Next, we can perform synthetic division of the polynomial by \( x - 1 \):

\[
\begin{array}{r|rrr}
1 & 1 & -6 & 11 & -6 \\
& & 1 & -5 & 6 \\
\hline
& 1 & -5 & 6 & 0 \\
\end{array}
\]

The result is \( x^2 - 5x + 6 \). Thus, we have:
\[
x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6)
\]

Now we can factor \( x^2 - 5x + 6 \):
\[
x^2 - 5x + 6 = (x - 2)(x - 3)
\]

Putting it all together, we can factor the original polynomial as:
\[
x^3 - 6x^2 + 11x - 6 = (x - 1)(x - 2)(x - 3)
\]

Thus, the additional linear factors of the polynomial \( x^3 - 6x^2 + 11x - 6 \) are \( x - 2 \) and \( x - 3 \).