Problem 107 - College Algebra

Question

Perform the indicated operations. $$\left(1-\frac{1}{x}\right)\left(1-\frac{1}{x+1}\right)\left(1-\frac{1}{x+2}\right)\left(1-\frac{1}{x+3}\right)$$

Step-by-Step Solution

Answer

The simplified form of \(\left(1-\frac{1}{x}\right)\left(1-\frac{1}{x+1}\right)\left(1-\frac{1}{x+2}\right)\left(1-\frac{1}{x+3}\right)\) is \(\frac{x^4 + 4x^3 + 6x^2}{(x+1)(x+2)(x+3)x}\).

Step 1: Distribute the Multiplication Operation

Take two pairs of factors and distribute respectively: \[\left(1-\frac{1}{x}\right)\left(1-\frac{1}{x+1}\right) = 1 - \frac{1}{x} - \frac{1}{x+1} + \frac{1}{x(x+1)}\]; \[\left(1-\frac{1}{x+2}\right)\left(1-\frac{1}{x+3}\right) = 1 - \frac{1}{x+2} - \frac{1}{x+3} + \frac{1}{(x+2)(x+3)}\]

Step 2: Combine Like Terms

Combine like terms of each expanded form: \[1 - \frac{1}{x} - \frac{1}{x+1} + \frac{1}{x(x+1)} = \frac{x(x+1) - x - (x+1) + 1}{x(x+1)} = \frac{x^2}{x(x+1)}\]; \[1 - \frac{1}{x+2} - \frac{1}{x+3} + \frac{1}{(x+2)(x+3)} = \frac{(x+2)(x+3) - (x+2) - (x+3) + 1}{(x+2)(x+3)} = \frac{x^2 + 4x + 6}{(x+2)(x+3)}\]

Step 3: Multiply the Results

Multiply the two equations derived from Steps 1 and 2: \[\frac{x^2}{x(x+1)} * \frac{x^2 + 4x + 6}{(x+2)(x+3)} = \frac{x^4 + 4x^3 + 6x^2}{(x+1)(x+2)(x+3)x}\]