Algebra 2

First, find a common denominator.

$\frac { 3 }{ { x }^{ 2 }+3x+2 } =\frac { A(x+1)+B(x+2) }{ (x+1)(x+2) }$

Thus,

$\frac { 3 }{ { x }^{ 2 }+3x+2 } =\frac { Ax+A+Bx+2B }{ { x }^{ 2 }+3x+2 }$

We can multiply both sides by

${ x }^{ 2 }+3x+2$

to get

$3\quad =\quad Ax+A+Bx+2B$

and rearrange to get

$3\quad =\quad Ax + Bx+A + 2B$

and then

$3\quad =\quad (A + B)x+(A + 2B)$

Since there is a $0x$ on the left side, then

$0\quad =\quad (A + B)x$

So

$A\quad +\quad B\quad =\quad 0$