Expand the brackets first Concept a(b+c)=a×b+a×c a(b−c)=a×b−a×c 3(5a2+2c)=3×5a2+3×2c 3(5a2+2c)=15a2+6c −2a(1−3a)=−2a×1−2a×−3a
NOTE: - x−=+ and a×a=a2 −2a(1−3a)=−2a+6a2 3(5a2+2c)−2a(1−3a)−6c=15a2+6c−2a+6a2−6c Group like terms 3(5a2+2c)−2a(1−3a)−6c=15a2+6a2−2a+6c−6c 3(5a2+2c)−2a(1−3a)−6c=21a2−2a+0 3(5a2+2c)−2a(1−3a)−6c=21a2−2a