## Precalculus: Mathematics for Calculus, 7th Edition

$\frac{4}{x}$
$Factor$ $the$ $expression$ $completely:$ $(1 + \frac{1}{x})^2 - (1 - \frac{1}{x})^2$ Square of a Sum Formula: $(A+B)^2 = A^2 + 2AB + B^2$ Square of a Difference Formula: $(A-B)^2 = A^2 - 2AB + B^2$ Use the Square of a Sum and Square of a Difference Formulas for both terms respectively $(1 + \frac{1}{x})^2$ = $1^2 + (2\times 1 \times \frac{1}{x}) + (\frac{1}{x})^2$ $(1 - \frac{1}{x})^2$ = $1^2 - (2\times 1 \times \frac{1}{x}) + (\frac{1}{x})^2$ $= (1+ \frac{2}{x} + \frac{1}{x^2}) - (1 - \frac{2}{x} + \frac{1}{x^2})$ Distribute the minus sign (-1) to $(1 - \frac{2}{x} + \frac{1}{x^2})$ and simplify $(1+ \frac{2}{x} + \frac{1}{x^2}) + (-1 + \frac{2}{x} - \frac{1}{x^2})$ $= 1+ \frac{2}{x} + \frac{1}{x^2} -1 + \frac{2}{x} - \frac{1}{x^2}$ $= \frac{4}{x}$