Resolve into factors :
`( x - 1) ( x + 1) ( x + 3) ( x + 5) + 7 `

Options

A. ` ( x + 2 + sqrt( 2)) ( x + 2 - 2 sqrt( 2)) ( x + 2 + sqrt(2)) ( x + 2 - 2 sqrt(2))`
B. ` ( x - 2 + sqrt(2)) ( x - 2 - sqrt(2))( x - 2 + 2 sqrt(2)) ( x + 2 - 2 sqrt(2))`
C. ` ( x - 2 - sqrt(2)) ( x + 2 + sqrt(2))( x - 2 - 2 sqrt(2)) ( x - 2 - 2 sqrt(2))`
D. none of these

Answer

To resolve the expression \((x - 1)(x + 1)(x + 3)(x + 5) + 7\) into factors, we will follow these steps:

### Step 1: Expand the expression
First, we will expand the product \((x - 1)(x + 1)\) and \((x + 3)(x + 5)\).

\[ (x - 1)(x + 1) = x^2 - 1 \]

\[ (x + 3)(x + 5) = x^2 + 8x + 15 \]

### Step 2: Combine the results
Now, we combine the results from Step 1:

\[ (x^2 - 1)(x^2 + 8x + 15) + 7 \]

### Step 3: Expand the combined expression
Next, we will expand \((x^2 - 1)(x^2 + 8x + 15)\):

\[ = x^2(x^2 + 8x + 15) - 1(x^2 + 8x + 15) \]

\[ = x^4 + 8x^3 + 15x^2 - x^2 - 8x - 15 \]

\[ = x^4 + 8x^3 + 14x^2 - 8x - 15 \]

Now, we add 7:

\[ = x^4 + 8x^3 + 14x^2 - 8x - 15 + 7 \]

\[ = x^4 + 8x^3 + 14x^2 - 8x - 8 \]

### Step 4: Factor the polynomial
Now we will factor the polynomial \(x^4 + 8x^3 + 14x^2 - 8x - 8\). We can use substitution to simplify our work. Let \(t = x^2 + 4x\):

\[ = (t^2 - 2t - 15) \]

### Step 5: Factor the quadratic
Now we can factor \(t^2 - 2t - 15\):

\[ = (t - 5)(t + 3) \]

### Step 6: Substitute back for \(t\)
Now we substitute back \(t = x^2 + 4x\):

\[ = (x^2 + 4x - 5)(x^2 + 4x + 3) \]

### Step 7: Factor each quadratic
Now we will factor each quadratic:

1. For \(x^2 + 4x - 5\):
- Roots are found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{16 + 20}}{2} = \frac{-4 \pm \sqrt{36}}{2} = \frac{-4 \pm 6}{2} \]
- Roots are \(1\) and \(-5\), so:
\[ = (x - 1)(x + 5) \]

2. For \(x^2 + 4x + 3\):
- Roots are:
\[ x = \frac{-4 \pm \sqrt{16 - 12}}{2} = \frac{-4 \pm 2}{2} \]
- Roots are \(-1\) and \(-3\), so:
\[ = (x + 1)(x + 3) \]

### Final Factorization
Combining all factors, we have:

\[ (x - 1)(x + 5)(x + 1)(x + 3) \]

### Conclusion
Thus, the expression \((x - 1)(x + 1)(x + 3)(x + 5) + 7\) factors to:

\[ (x - 1)(x + 5)(x + 1)(x + 3) \]