Question

Factorise 6x^2+7ab+2b^2+11a+7b+4/3​

Answer 1

Answer:

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

Step-by-step explanation:

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)

Answer 2

Step-by-step explanation:

i hardly tried..to this..