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The given expression is:

\[
\left( x^2 + 8x \right)^2 - 2\left( x^2 + 8x \right) - 63
\]

### Step 1: Substitution to simplify

Let’s substitute \( y = x^2 + 8x \). This transforms the expression into a quadratic form in terms of \( y \):

\[
y^2 - 2y - 63
\]

### Step 2: Factor the quadratic

Now, factor the quadratic expression:

\[
y^2 - 2y - 63 = (y - 9)(y + 7)
\]

### Step 3: Substitute back \( y = x^2 + 8x \)

Substitute \( y = x^2 + 8x \) back into the factored form:

\[
(x^2 + 8x - 9)(x^2 + 8x + 7)
\]

### Step 4: Factor each quadratic expression

Now, we factor each quadratic:

1. **Factor \( x^2 + 8x - 9 \):**

   Find factors of \(-9\) that add up to \(8\). These are \(9\) and \(-1\).

   \[
   x^2 + 8x - 9 = (x + 9)(x - 1)
   \]

2. **Factor \( x^2 + 8x + 7 \):**

   Find factors of \(7\) that add up to \(8\). These are \(7\) and \(1\).

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

### Step 5: Write the final factored form

The expression is now factored as:

\[
(x + 9)(x - 1)(x + 7)(x + 1)
\]

This is the product of four linear factors.

---

Would you like any more details or clarifications on this process?

Here are 5 related questions for further exploration:
1. How can we check if the factors are correct by expanding the product?
2. How does completing the square apply to factoring quadratics?
3. What are other methods to factor quadratic expressions?
4. How do the properties of polynomial roots relate to linear factors?
5. Can this approach be extended to higher-degree polynomials?

**Tip**: When factoring quadratics, always try substitution if the form resembles a perfect square or simple quadratic pattern.

Rewrite the expression as a product of four linear factors: (x^2 + 8x)^2 - 2(x^2 + 8x) - 63

Learn how to simplify and factor the quadratic expression (x^2 + 8x)^2 - 2(x^2 + 8x) - 63. This problem covers key algebraic concepts including factoring, substitution, and solving quadratic equations. Suitable for students in Grades 9-11.

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