(x+2)(x+3)(x+4)(x+5)-8

4 min read Jun 16, 2024

Factoring the expression (x+2)(x+3)(x+4)(x+5)-8

The expression (x+2)(x+3)(x+4)(x+5)-8 might look intimidating at first glance, but we can factor it in a clever way to reveal its hidden beauty.

Step 1: Recognizing the Pattern

Observe that the first part of the expression is a product of four consecutive terms. This hints at a pattern that can be exploited for factorization.

Step 2: Introducing a "Middle Term"

Let's introduce a new variable, say 'y', to simplify our expression. Let:

y = (x+2)(x+5)

Notice that if we multiply out the terms of (x+3)(x+4), we get:

(x+3)(x+4) = x² + 7x + 12

Now, we can rewrite the original expression in terms of 'y':

(x+2)(x+3)(x+4)(x+5) - 8 = y(x² + 7x + 12) - 8

Step 3: Completing the Square

Let's focus on the quadratic term (x² + 7x + 12). We can manipulate it by completing the square. To do this:

Take half of the coefficient of the x term (7/2) and square it (49/4).
Add and subtract this value inside the parentheses:

(x² + 7x + 12) = (x² + 7x + 49/4 - 49/4 + 12)

The first three terms now form a perfect square trinomial:

(x² + 7x + 49/4 - 49/4 + 12) = (x + 7/2)² - 1/4

Step 4: Substituting and Factoring

Substitute this back into our expression:

y(x² + 7x + 12) - 8 = y((x + 7/2)² - 1/4) - 8

Now, we can factor out a common factor of 'y':

y((x + 7/2)² - 1/4) - 8 = y(x + 7/2)² - y/4 - 8

We can rewrite the last two terms to get a difference of squares:

y(x + 7/2)² - y/4 - 8 = y(x + 7/2)² - (1/4)(y + 32)

Finally, we can factor this expression as a difference of squares:

y(x + 7/2)² - (1/4)(y + 32) = [√(y)(x + 7/2) + √(1/4)(y + 32)][√(y)(x + 7/2) - √(1/4)(y + 32)]

Step 5: Replacing 'y'

Remember that y = (x+2)(x+5). Substitute this back into the factored expression to get the final result:

[√((x+2)(x+5))(x + 7/2) + √(1/4)((x+2)(x+5) + 32)][√((x+2)(x+5))(x + 7/2) - √(1/4)((x+2)(x+5) + 32)]

Conclusion

This factorization might seem intricate, but it demonstrates the power of recognizing patterns, completing the square, and utilizing variable substitutions to simplify complex expressions.