Question 862244
{{{x^2-5x+6=(x-2)(x-3)}}}
{{{x^2+2x-15=(x+5)(x-3)}}}
{{{(x^2-5x+6)/(x^2+2x-15)=((x-2)(x-3))/((x+5)(x-3))}}}
{{{(x^2-5x+6)/(x^2+2x-15)=(x-2)/(x+5)}}}
Since there is a removable discontinuity at {{{x=3}}}, the only restriction on the domain is {{{x=-5}}}