Algebra: A Combined Approach (4th Edition)

$\dfrac{x^{2}+xa+xb+ab}{x^{2}-xc+bx-bc}=\dfrac{x+a}{x-c}$
$\dfrac{x^{2}+xa+xb+ab}{x^{2}-xc+bx-bc}$ Group the first two terms and the last two terms in the numerator together. Do the same in the denominator: $\dfrac{x^{2}+xa+xb+ab}{x^{2}-xc+bx-bc}=\dfrac{(x^{2}+xa)+(xb+ab)}{(x^{2}-xc)+(bx-bc)}=...$ Take out common factor $x$ from the first parentheses in both the numerator and the denominator and common factor $b$ from the second parentheses in both the numerator and the denominator: $...=\dfrac{x(x+a)+b(x+a)}{x(x-c)+b(x-c)}=...$ Take out common factor $x+a$ from the numerator and common factor $x-c$ from the denominator and simplify: $...=\dfrac{(x+a)(x+b)}{(x+b)(x-c)}=\dfrac{x+a}{x-c}$