Answer
$\dfrac{x^{2}+xa+xb+ab}{x^{2}-xc+bx-bc}=\dfrac{x+a}{x-c}$
Work Step by Step
$\dfrac{x^{2}+xa+xb+ab}{x^{2}-xc+bx-bc}$
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:
$...=\dfrac{(x+a)(x+b)}{(x+b)(x-c)}=\dfrac{x+a}{x-c}$
