## Intermediate Algebra (6th Edition)

$24a^{3} +28a^{2}b-6a-7b$
$(2a+1)(2a-1)(6a+7b)$ Multiplying $(2a+1)(2a-1),$ it is in the form of $(a+b)(a-b)$ $(a+b)(a-b)= (a^{2}-b^{2})$. $(2a+1)(2a-1)=(2a)^{2}-1^{2} = 4a^{2} - 1$ $(2a+1)(2a-1)(6a+7b) = (4a^{2} - 1)(6a+7b)$ $=24a^{3}+28a^{2}b-6a-7b$