關於中2的數學題

Method I:

(2p+3q)^2 - (2p-3q)^2
= [(2p+3q) + (2p-3q)][(2p+3q) - (2p-3q)]
= [4p][6q]
= 24pq

Method II:

(2p+3q)^2 - (2p-3q)^2
= [(2p)^2 + 2(2p)(3q) + (3q)^2] - [(2p)^2 - 2(2p)(3q) + (3q)^2]
= [4p^2 + 12pq + 9q^2] - [4p^2 - 12pq + 9q^2]
= 24pq

do you know there are some perfect square identity?

for this, you should use this: (a+b)(a-b) = a^2 - b^2

if you know that, you would know how to do it right a way

(2p+3q)^2-(2p-3q)^2
= [(2p+3q)+(2p-3q)][(2p+3q)-(2p-3q)]
= (4p)(6q)
=24pq

apart from this, there are two simliar identities, you better remember them

1. (a+b)^2 = a^2 + 2ab + b^2
2. (a-b)^2 = a^2 - 2ab + b^2

(2p+3q)2-(2p-3q)2
A2 - B2 =( A+B )( A-B )
=( 2p+3q+2p-3q )( 2p+3q-2p+3q )
=( 4p )( 6q )
=24pq