We have,
a2−(2a+3b)2=(a−(2a+3b))(a+(2a+3b))∴a2−b2=(a+b)(a−b)
=(a−2a−3b)(a+2a+3b)
=(−a−3b)(3a+3b)
=−3(a+3b)(a+b)
Hence, this is the answer.
Solve the following equations by using formula
(1) a2 − 2a − 4 = 0
(2) x2 − 8x + 1 = 0
(3) m2 − 2m − 2 = 0
(4) k2 − 6k = 1
(5) 2y2 + 6y = 3
(6) 8r2 = r + 2
(7) p = 5 − 2p2
(8) 2z2 + 7z = 4 = 0
(9) 3b2 + 2b = 2
(10) a2 = 4a + 6