Given:#" "(b/(b+3)+2)/(b^2-2b-8 )#
#color(blue)("Consider the numerator: "b/(b+3)+2/1)#
I have deliberately chosen to write 2 as #2/1#. Although correct this is not normally done.
#color(green)([b/(b+3)]+[2/1color(red)(xx1)])#
#color(green)([b/(b+3)]+[2/1color(red)(xx(b+3)/(b+3))])#
#[b/(b+3)]+[(2b+6)/(b+3)]#
#(3b+6)/(b+3)#
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#color(blue)("Consider the denominator: "b^2-2b-8)#
Notice that #2xx(-4)=-8 and 2-4=-2#
Write as #(b-4)(b+2)#
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#color(blue)("Putting it all together")#
numerator#-:# denominator#->(3b+6)/(b+3)-:(b-4)(b+2)#
#" "=" "(3b+6)/(b+3)xx1/((b-4)(b+2))#
Factor out the 3 from #3b+6->3(b+2)#
#" "=" "(3cancel((b+2)))/(b+3)xx1/((b-4)cancel((b+2)))#
#" "=" "3/(b^2-b-12)#
