6a^2+2b^2+3c^2+7ab+11ac+7bc

Question

the above maths should be simplification.

Answer 1

6a^2+2b^2+3c^2+7ab+7bc+11ac

=2b^2+7b(a+c)+(6a^2+11ac+3c^2)

=2b^2+7b(a+c)+(6a^2+9ac+2ac+3c^2)

=2b^2+7b(a+c)+{3a(2a+3c)+c(2a+3c)}

=2b^2+7b(a+c)+(2a+3c)(3a+c)

=2b^2+{(4a+6c)+(3a+c)}b+(2a+3c)(3a+c)

=2b^2+(4a+6c)b+(3a+c)b+(2a+3c)(3a+c)

=2b(b+2a+3c)+(3a+c)(b+2a+3c)

=(b+2a+3c)(2b+3a+c)

By Arindam Kundu SAVM Chandannagar

Answer 2

Another Process 6a^2+2b^2+3c^2+7bc+7ab+11ac On Spliting We get, 6a^2+4ab+2ac+2b^2+bc+3ab+3c^2+6bc+9ac =2a(c+2b+3a)+b(c+2b+3a)+3c(c+2b+3a) =(c+2b+3a)(3c+b+2a) By Safal Das Biswas S.A.V.M(Chinsurah)