We have a rectangle with length L=x-8 and width W=x+11
Let's remember that the area of a rectangle is given by the following equation
[tex]A_R=L\cdot W[/tex]We replace these values
[tex]A_R=(x+11)(x-8)[/tex]We multiply and solve
[tex]\begin{gathered} A_R=x^2-8x+11x-88 \\ A_R=x^2+3x-88 \end{gathered}[/tex]The answer is option B