Answer
$a=5$
Work Step by Step
$\frac{4}{a^2+4a+3}+\frac{2}{a^2+a-6}-\frac{3}{a^2-a-2}=0$
$\frac{4}{(a+1)(a+3)}+\frac{2}{(a+3)(a-2)}-\frac{3}{(a-2)(a+1)}=0$
$[(a+1)(a+3)(a-2)](\frac{4}{(a+1)(a+3)}+\frac{2}{(a+3)(a-2)}-\frac{3}{(a-2)(a+1)}=0)$
$4(a-2)+2(a+1)-3(a+3)=0(a+1)(a+3)(a-2)$
$4a-8+2a+2-3a-9=0$
$3a-15=0$
$3a=15$
$3a/3=15/3$
$a=5$
$\frac{4}{a^2+4a+3}+\frac{2}{a^2+a-6}-\frac{3}{a^2-a-2}=0$
$\frac{4}{5^2+4*5+3}+\frac{2}{5^2+5-6}-\frac{3}{5^2-5-2}=0$
$\frac{4}{25+20+3}+\frac{2}{25+5-6}-\frac{3}{25-5-2}=0$
$\frac{4}{48}+\frac{2}{24}-\frac{3}{18}=0$
$\frac{1}{12}+\frac{1}{12}-\frac{1}{6}=0$
$\frac{1}{12}+\frac{1}{12}-\frac{2}{12}=0$
$\frac{2}{12}-\frac{2}{12}=0$
