b) (2x – 1)2 – (x + 3)2 = 0
=> (2x – 1 – x – 3)(2x – 1 + x + 3) = 0
=> (x – 4)(3x + 2) = 0
=> x – 4 = 0 hay 3x + 2 = 0
c) x2(x – 3) + 12 – 4x = 0
=> x2(x – 3) – 4(x – 3) = 0
=> (x – 3)(x2 – 4) = 0
=> (x – 3)(x – 2)(x + 2) = 0
=> x – 3 = 0 ; x – 2 = 0 hay x + 2 = 0
=> x = 3 ; x = 2 hay x = -2