I keep looking and writing out the problem, but I don't know where $(y^2 + 1)$ comes from. I get $y^2(y-3)$ as the numerator instead, but I can't tell if I'm having a brain fart and am missing something.
$\endgroup$ 31 Answer
$\begingroup$$$y^2(y-3)+(y-3)=y^2(y-3)+(1)(y-3)=(y^2+1)(y-3)$$
We just factorize $y-3$ out.
$\endgroup$ 1