Question

What are the remaining linear factors of x4−13x3−x2+13x with known factors (x+1) and (x−1) ?(1 point) Responses (x−1) and (x−13) left parenthesis x minus 1 right parenthesis and left parenthesis x minus 13 right parenthesis (x2−13) left parenthesis x squared minus 13 right parenthesis (x2−13x) left parenthesis x squared minus 13 x right parenthesis x and (x−13)

Answer 1

To find the remaining linear factors of the polynomial \( x^4 - 13x^3 - x^2 + 13x \) given the factors \( (x + 1) \) and \( (x - 1) \), we can start by performing polynomial long division or synthetic division.

### Step 1: Checking the known factors

Let's check if \( (x + 1) \) and \( (x - 1) \) are indeed factors by substituting \( x = -1 \) and \( x = 1 \) into the polynomial.

1. For \( x = -1 \):
\[
(-1)^4 - 13(-1)^3 - (-1)^2 + 13(-1) = 1 + 13 - 1 - 13 = 0
\]
So \( x + 1 \) is a factor.

2. For \( x = 1 \):
\[
(1)^4 - 13(1)^3 - (1)^2 + 13(1) = 1 - 13 - 1 + 13 = 0
\]
So \( x - 1 \) is also a factor.

### Step 2: Polynomial division

We'll now divide \( x^4 - 13x^3 - x^2 + 13x \) by \( (x + 1)(x - 1) = x^2 - 1 \).

**Perform the division:**

1. Divide \( x^4 \) by \( x^2 \) to get \( x^2 \).
2. Multiply \( x^2 \) by \( x^2 - 1 \) to get \( x^4 - x^2 \).
3. Subtract:
\[
(x^4 - 13x^3 - x^2 + 13x) - (x^4 - x^2) = -13x^3 + 13x
\]
4. Now divide \(-13x^3\) by \(x^2\) to get \(-13x\).
5. Multiply \(-13x\) by \(x^2 - 1\) to get \(-13x^3 + 13x\).
6. Subtract:
\[
(-13x^3 + 13x) - (-13x^3 + 13x) = 0
\]

The polynomial factors as follows:
\[
x^4 - 13x^3 - x^2 + 13x = (x^2 - 1)(x^2 - 13x)
\]

### Step 3: Factoring further

Now we need to factor \( x^2 - 13x \):
\[
x^2 - 13x = x(x - 13)
\]

### Step 4: Complete Factorization

Putting it all together, we have:
\[
x^4 - 13x^3 - x^2 + 13x = (x + 1)(x - 1)(x)(x - 13)
\]

### Final Answer

The remaining linear factors are:
- \( x \)
- \( (x - 13) \)

So the remaining linear factors are \( x \) and \( (x - 13) \).