Question

What are the additional linear factors of x3−3x2−6x+8
if x−1
is a factor?



(1 point)
Responses

(x+4)(x−2)
open paren x plus 4 close paren times open paren x minus 2 close paren

(x−4)(x−2)
open paren x minus 4 close paren times open paren x minus 2 close paren

(x−4)(x+2)
open paren x minus 4 close paren times open paren x plus 2 close paren

(x+4)(x+2)
open paren x plus 4 close paren times open paren x plus 2 close paren
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page 14 of 14

Answer 1

To determine the additional linear factors of the polynomial \( x^3 - 3x^2 - 6x + 8 \) given that \( x - 1 \) is a factor, we can perform polynomial division or synthetic division of \( x^3 - 3x^2 - 6x + 8 \) by \( x - 1 \).

### Step 1: Perform Synthetic Division

1. Set up the synthetic division with the root \( x = 1 \):

\[
\begin{array}{r|rrrr}
1 & 1 & -3 & -6 & 8 \\
& & 1 & -2 & -8 \\
\hline
& 1 & -2 & -8 & 0 \\
\end{array}
\]

### Step 2: Write down the results

The bottom row indicates the coefficients of the quotient polynomial, which is:

\[
1x^2 - 2x - 8 = x^2 - 2x - 8
\]

### Step 3: Factor the quotient polynomial \( x^2 - 2x - 8 \)

Next, we need to factor the quadratic \( x^2 - 2x - 8 \). We are looking for two numbers that multiply to \(-8\) (the constant term) and add to \(-2\) (the coefficient of \(x\)). The correct pair is \(-4\) and \(2\).

Thus:

\[
x^2 - 2x - 8 = (x - 4)(x + 2)
\]

### Step 4: Write the complete factorization

Given that \( x - 1 \) is a factor, the complete factorization of the polynomial \( x^3 - 3x^2 - 6x + 8 \) is:

\[
(x - 1)(x - 4)(x + 2)
\]

### Conclusion

The additional linear factors of the polynomial, excluding \( x - 1 \), are:

\[
(x - 4)(x + 2)
\]

Therefore, the correct response is:

\((x - 4)(x + 2)\)
open paren x minus 4 close paren times open paren x plus 2 close paren.