Explanation:
Here we have the following polynomial:
[tex]6v(2v+3)[/tex]
The problem states that this expression represents an area model so we can think of this as the area of a rectangle whose dimensions are:
[tex]L=6v \\ \\ W=2v+3 \\ \\ \\ L:Length \\ \\ W:Width \\ \\ \\ A=L\times W \\ \\ A=6v(2v+3) \\ \\ \\ A:Area[/tex]
Therefore, by applying distributive property:
[tex]6v(2v+3)=6v(2v)+6v(3)=(6)(2)v\cdot v+(6)(3)v=\boxed{12v^2+18v}[/tex]
