Factoring and Solving the Expression (x+1)(x+9)(x+5)^2 + 63
This article explores the process of factoring and solving the expression (x+1)(x+9)(x+5)^2 + 63.
Understanding the Expression
The given expression is a polynomial with a degree of 5, meaning it has five possible roots. It's a combination of linear and squared factors, making it potentially challenging to factor directly.
Strategies for Factoring

Expansion and Grouping:
 Expand the expression to obtain a standard polynomial form.
 Attempt to group terms and factor by grouping. However, this approach can be quite tedious and may not lead to a successful factorization.

Substitution:
 Consider a substitution to simplify the expression.
 Let y = (x+5).
 The expression becomes (y4)(y+4)y^2 + 63, which simplifies to y^4  16y^2 + 63.
 This new expression is a quadratic in y^2. We can factor it as (y^2  7)(y^2  9).
 Substitute back y = (x+5) to get ((x+5)^2  7)((x+5)^2  9).

Recognizing Patterns:
 Observe that the expression resembles the expansion of a perfect square.
 Notice that ((x+5)^2  7)((x+5)^2  9) can be rewritten as ((x+5)^2  8)^2  1.
 This pattern allows us to factor the expression further using the difference of squares formula: (a^2  b^2) = (a+b)(ab).
 The final factored expression becomes ((x+5)^2  8 + 1)((x+5)^2  8  1), which simplifies to ((x+5)^2  7)((x+5)^2  9).
Solving the Equation
To find the roots (solutions) of the expression, we set it equal to zero:
((x+5)^2  7)((x+5)^2  9) = 0
This equation is satisfied when either factor equals zero:
 (x+5)^2  7 = 0
 (x+5)^2  9 = 0
Solving for x in each case:

(x+5)^2 = 7

x + 5 = ±√7

x = 5 ±√7

(x+5)^2 = 9

x + 5 = ±√9

x = 5 ±3
Therefore, the solutions to the equation are:
 x = 5 + √7
 x = 5  √7
 x = 2
 x = 8
Conclusion
Factoring and solving the expression (x+1)(x+9)(x+5)^2 + 63 involves recognizing patterns, using substitution, and applying factoring techniques. The solutions, or roots, of the equation are x = 5 + √7, x = 5  √7, x = 2, and x = 8. This demonstrates the power of algebraic manipulations and pattern recognition in simplifying complex expressions.