## Algebra: A Combined Approach (4th Edition)

$A=(4x^{2}-9)$ square inches
The top of the table is a rectangle, so its area $(A)$ will be $A= L \times W,$ where $L=(2x+3)$ inches and $W=(2x-3)$ inches. Therefore, $A=(2x+3)(2x-3)$ $A=2x(2x-3)+3(2x-3)$ $A=4x^{2}-6x+6x-9$ $A=4x^{2}-9$ Therefore, the area of the table is $A=(4x^{2}-9)$ square inches.