ANSWER
The restrictions are
[tex]a\ne -3,a\ne -\frac{1}{5}[/tex]
EXPLANATION
We were given the rational function,
[tex]\frac{2a^2+a-15}{5a^2+16a+3}[/tex]
The function is defined for all values of a, except
[tex]5a^2+16a+3=0[/tex]
This has become a quadratic trinomial, so we need to split the middle term.
We do that by multiplying the coefficient of [tex]x^2[/tex] which is 5 by the constant term which is 3. This gives us 15.
The factors of 15 that adds up to 16 are 1 and 15.
We use these factors to split the middle term.
[tex]5a^2+15a+a+3=0[/tex]
We now factor to get,
[tex]5a(a+3)+1(a+3)=0[/tex]
We factor further to get,
[tex](a+3)(5a+1)=0[/tex]
This implies that,
[tex](a+3)=0,(5a+1)=0[/tex]
This gives
[tex]a=-3,a=-\frac{1}{5}[/tex]
These are the restrictions.
