Question 117515
{{{(6y/(y^2+5y+4))+(4y/(y^2-1))}}}


Factoring both denominators is the key to doing this problem just as you suspected.  Since 4 X 1 = 4 and 4 + 1 = 5, this trinomial factors rather neatly.


{{{y^2+5y+4=(y+4)(y+1)}}}.


Now, you can write your problem with fully factored denominators, thus:


{{{(6y/((y+4)(y+1)))+(4y/((y-1)(y+1)))}}}


Next you need a common denominator.  In this case, you have one factor of {{{y+4}}}, one of {{{y+1}}}, and one of {{{y-1}}}, so:


{{{(6y(y-1)+4y(y+4))/((y+4)(y-1)(y+1)))}}}
{{{(6y^2-6y+4y^2+16y)/((y+4)(y-1)(y+1)))}}}
{{{(10y^2+10y)/((y+4)(y-1)(y+1)))}}}


Next, factor the numerator


{{{(10y(y+1)/((y+4)(y-1)(y+1)))}}}


Eliminate {{{(y+1)/(y+1)}}}


{{{(10y/((y+4)(y-1)))}}}


Sometimes it might be more convenient to leave your answer just like that.  At least the domain of the function is readily apparent (all Reals except -4 and 1).  But if need be, multiply the binomial, thus:


{{{(10y/(y^2+3y-4)))}}}


Hope that helps,
John