The factor form `z^(2)+(1)/(z^(2))+2-2z-(2)/(z)` is

Options

A. `(z+(1)/(z)+2) (z-(1)/(z))`
B. `(z+(1)/(z))(z+(1)/(z)-2)`
C. `(z-(1)/(z)+2)(z+(1)/(z))`
D. `(z-(1)/(z))(z-(1)/(z)-2)`

Answer

To factor the expression \( z^2 + \frac{1}{z^2} + 2 - 2z - \frac{2}{z} \), we can follow these steps:

### Step 1: Rewrite the expression
Start with the original expression:
\[ z^2 + \frac{1}{z^2} + 2 - 2z - \frac{2}{z} \]

### Step 2: Recognize patterns
Notice that \( z^2 + \frac{1}{z^2} + 2 \) can be rewritten using the identity:
\[ a^2 + b^2 + 2ab = (a + b)^2 \]
where \( a = z \) and \( b = \frac{1}{z} \). Thus,
\[ z^2 + \frac{1}{z^2} + 2 = \left(z + \frac{1}{z}\right)^2 \]

### Step 3: Substitute back into the expression
Substituting this back into the expression gives:
\[ \left(z + \frac{1}{z}\right)^2 - 2z - \frac{2}{z} \]

### Step 4: Rewrite \( -2z - \frac{2}{z} \)
Notice that \( -2z - \frac{2}{z} \) can also be expressed in terms of \( z + \frac{1}{z} \):
\[ -2(z + \frac{1}{z}) \]

### Step 5: Combine the terms
Now the expression can be rewritten as:
\[ \left(z + \frac{1}{z}\right)^2 - 2\left(z + \frac{1}{z}\right) \]

### Step 6: Factor out the common term
Let \( x = z + \frac{1}{z} \). Then, the expression simplifies to:
\[ x^2 - 2x \]
This can be factored as:
\[ x(x - 2) \]

### Step 7: Substitute back for \( x \)
Substituting back \( x = z + \frac{1}{z} \):
\[ (z + \frac{1}{z})(z + \frac{1}{z} - 2) \]

### Final Result
Thus, the factor form of the original expression is:
\[ (z + \frac{1}{z})(z + \frac{1}{z} - 2) \]