How does one expression factor into the other? How does $$(k+1)(k^2+2k)(3k+5)$$ factor into $$(k)(k^2-1)(3k+2) + 12k(k+1)^2$$

Well, $RHS=k(k^2-1)(3k+2) + 12k(k+1)^2 = k(k+1)((k-1)(3k+2)+12(k+1))=k(k+1)(3k^2+2k-3k-2+12k+12)=k(k+1)(3k^2+11k+10)=k(k+1)(k+2)(3k+5)=LHS$
