Factorise:

We have:

$(2 x+5 y)^{2}-1=(2 x+5 y)^{2}-(1)^{2}$

$=\{(2 x+5 y)+1\}\{(2 x+5 y)-1\}$

$=(2 x+5 y+1)(2 x+5 y-1)$

$\therefore(2 x+5 y)^{2}-1=(2 x+5 y+1)(2 x+5 y-1)$