=3k(2k²)-3k(3k)-3k(5)+4(2k²)-4(3k)-4(5)
=6k³-9k²-15k+8k²-12k-20
=6k³-k²-27k-20
....
Break it up into four individual rectangles' areas.
Two have length of x and width of 1, two have length of x-2 (because those 1 width/length squares in the corners are covered by the full length rectangles) and width of 1.
So total area
=x•1+x•1+(x-2)1+(x-2)1
=x+x+x-2+x-2
=4x-4
