Question

What are the additional linear factors of x3−3x2−6x+8
𝑥
3
−
3
𝑥
2
−
6
𝑥
+
8
if x−1
𝑥
−
1
is a factor?



(1 point)
Responses

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

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

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

(x+4)(x+2)

Answer 1

Given that \( x - 1 \) is a factor of the polynomial \( x^3 - 3x^2 - 6x + 8 \), we can use polynomial long division to divide the polynomial by \( x - 1 \) to find the other factors.

Performing polynomial long division:

1. Divide the leading term of the dividend \( x^3 \) by the leading term of the divisor \( x \) to get \( x^2 \).
2. Multiply \( x^2 \) by \( x - 1 \) to get \( x^3 - x^2 \).
3. Subtract this from the original polynomial:

\[
(x^3 - 3x^2 - 6x + 8) - (x^3 - x^2) = -2x^2 - 6x + 8
\]

4. Now divide the leading term \( -2x^2 \) by \( x \) to get \( -2x \).
5. Multiply \( -2x \) by \( x - 1 \) to get \( -2x^2 + 2x \).
6. Subtract this:

\[
(-2x^2 - 6x + 8) - (-2x^2 + 2x) = -8x + 8
\]

7. Now divide \( -8x \) by \( x \) to get \( -8 \).
8. Multiply \( -8 \) by \( x - 1 \) to get \( -8x + 8 \).
9. Subtract this:

\[
(-8x + 8) - (-8x + 8) = 0
\]

Now we have completed the division and the quotient is \( x^2 - 2x - 8 \).

Next, we need to factor \( x^2 - 2x - 8 \). We can look for two numbers that multiply to \(-8\) and add to \(-2\). These two numbers are \( -4 \) and \( 2 \).

So, we factor \( x^2 - 2x - 8 \) as:

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

Thus, the complete factorization of the original polynomial is:

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

Therefore, the additional linear factors (given that \( x - 1 \) is a factor) are:

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

So the answer is:

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