# How do you combine (x+1)/(x^2+6x+8) + (x-4)/(x^2-3x-10)?

May 27, 2015

Convert the two terms so they have a common denominator then add the numerators (and possibly simplify)

Since ${x}^{2} + 6 x + 8 = \textcolor{red}{\left(x + 2\right)} \left(x + 4\right)$
and ${x}^{2} - 3 x - 10 = \textcolor{red}{\left(x + 2\right)} \left(x - 5\right)$
the obvious candidate for denominator is
$\left(x + 2\right) \left(x + 4\right) \left(x - 5\right)$

and
$\frac{x + 1}{{x}^{2} + 6 x + 8} + \frac{x - 4}{{x}^{2} - 3 x - 10}$

can be written as

$\frac{\left(x + 1\right) \left(x - 5\right) + \left(x - 4\right) \left(x + 4\right)}{\left(x + 2\right) \left(x + 4\right) \left(x - 5\right)}$

$= \frac{{x}^{2} - 4 x - 5 + {x}^{2} - 16}{\left(x + 2\right) \left(x + 4\right) \left(x - 5\right)}$

$= \frac{2 {x}^{2} - 4 x - 21}{\left(x + 2\right) \left(x + 4\right) \left(x - 5\right)}$

(with no obvious factors of the numerator)
= (2x^2-4x-21)/(x^3+x^2-22x-40