The correct option is C (3a−5)(a+1)(a−1)(3a+1)
5−(3a2−2a)(6−3a2+2a)
=5−6(3a2−2a)+(3a2−2a)2
=[(3a2−2a)2−5(3a2−2a)]−[(3a2−2a)−5]
=(3a2−2a)(3a2−2a−5)−1(3a2−2a−5)
=(3a2−2a−5)(3a2−2a−1)
=(3a2−5a+3a−5)(3a2−3a+a−1)
=[3a(a+1)−5(a+1)][3a(a−1)+(a−1)]
=(a+1)(3a−5)(a−1)(3a+1)