Question

Factorise: 64p^2 - (2q+5r)^2
Please show with steps please! ​

Answer 1

$64p {}^{2} - 1(4p {}^{2} + 10qr + 5r(2q + 5r))$

$64p {}^{2} - 1(4q {}^{2} + 10qr + 25r {}^{2} ))$

$64p {}^{2} - 1(2q + 5r) {}^{2}$

$64p {}^{2} - 1(2p + 5r)(2q + 5r)$

$64p {}^{2} - 1(4q {}^{2} + 20qr + 25r {}^{2} )$

$64p {}^{2} - 1(2p(2q + 5r) + (5r(2q + 5r))$

$64p {}^{2} - 4q {}^{2} - 20qr - 25r {}^{2}$

$solution$

$64p {}^{2} - 4q {}^{2} - 20qr - 25r {}^{2}$