The HCF of `x^6 - 1` and `x^4 + 2x^3 - 2x^(1) - 1` is:

Options

A. `x^(2)+1`
B. `x-1`
C. `x^(2)-1`
D. `x+1`

Answer

To find the HCF of the polynomials \(x^6 - 1\) and \(x^4 + 2x^3 - 2x - 1\), we can follow these steps:

### Step 1: Factor \(x^6 - 1\)
We start by factoring the polynomial \(x^6 - 1\). We can recognize that this can be expressed as a difference of cubes:
\[ x^6 - 1 = (x^2)^3 - 1^3 \]
Using the identity for the difference of cubes, \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\), we set \(a = x^2\) and \(b = 1\):
\[ x^6 - 1 = (x^2 - 1)(x^4 + x^2 + 1) \]
Next, we can further factor \(x^2 - 1\) as it is a difference of squares:
\[ x^2 - 1 = (x - 1)(x + 1) \]
Thus, we have:
\[ x^6 - 1 = (x - 1)(x + 1)(x^4 + x^2 + 1) \]

### Step 2: Factor \(x^4 + 2x^3 - 2x - 1\)
Next, we need to factor the second polynomial \(x^4 + 2x^3 - 2x - 1\). We can group the terms:
\[ x^4 + 2x^3 - 2x - 1 = (x^4 + 2x^3) + (-2x - 1) \]
Factoring out \(x^3\) from the first group:
\[ = x^3(x + 2) - 1(2x + 1) \]
This does not seem to lead us to a factorization directly. Instead, we can try polynomial long division or synthetic division to check for roots or factors.

### Step 3: Find common factors
To find the HCF, we can check if \(x^2 - 1\) is a factor of \(x^4 + 2x^3 - 2x - 1\). We know that \(x^2 - 1 = (x - 1)(x + 1)\). We will check if \(x = 1\) and \(x = -1\) are roots of \(x^4 + 2x^3 - 2x - 1\).

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

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

Since both \(x = 1\) and \(x = -1\) are roots, \(x^2 - 1\) is indeed a factor of \(x^4 + 2x^3 - 2x - 1\).

### Step 4: Conclusion
The common factor between \(x^6 - 1\) and \(x^4 + 2x^3 - 2x - 1\) is \(x^2 - 1\). Therefore, the HCF of the two polynomials is:
\[ \text{HCF} = x^2 - 1 \]