[tex]\displaystyle \frac{4n^2-9n+2}{2+9n-5n^2}=\frac{4n^2-n-8n+2}{-5n^2+9n+2}=\frac{n(4n-1)-2(4n-1)}{-5n^2+10n-n+2}=\frac{(4n-1)(n-2)}{-5n(n-2)-(n-2)}=\\\frac{(4n-1)(n-2)}{-(n-2)(5n+1)}=\frac{(4n-1)(-1)}{5n+1}=-\frac{4n-1}{5n+1}=\frac{1-4n}{5n+1}[/tex]