Question

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

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 or synthetic division of the cubic polynomial by \( x - 3 \).

Let's use synthetic division:

1. Write down the coefficients of the polynomial: \( 1, -6, 11, -6 \).
2. Use \( 3 \) (the root associated with the factor \( x - 3 \)) for synthetic division.

\[
\begin{array}{r|rrrr}
3 & 1 & -6 & 11 & -6 \\
& & 3 & -9 & 6 \\
\hline
& 1 & -3 & 2 & 0 \\
\end{array}
\]

The row at the bottom gives us the coefficients of the quotient polynomial: \( 1, -3, 2 \), which corresponds to:

\[
x^2 - 3x + 2
\]

Now, we can factor \( x^2 - 3x + 2 \):

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

Putting it all together, the original polynomial can be expressed as:

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

Thus, the additional linear factors are \( x - 1 \) and \( x - 2 \).