I would solve this differently. I'm using * for times.
Problem 1:
We have (1-x)e2 + (1-x)e3, which is (1-x)e2 + {(1-x)e2 * (1-x)}. Each term has (1-x)e2 in it, so I factor it out to get (1-x)e2 * {1 + (1-x)} or (1-x)e2 * (2-x). (Notice that Harv dropped a minus sign - F and A are opposites of each other.)
Problem 2:
(π - 4)e2 + π - 4 has n-4 in each term. Factor it out to get (n-4) * {(n-4) + 1}, so we have (n-4)* (n-3).
Problem 1:
We have (1-x)e2 + (1-x)e3, which is (1-x)e2 + {(1-x)e2 * (1-x)}. Each term has (1-x)e2 in it, so I factor it out to get (1-x)e2 * {1 + (1-x)} or (1-x)e2 * (2-x). (Notice that Harv dropped a minus sign - F and A are opposites of each other.)
Problem 2:
(π - 4)e2 + π - 4 has n-4 in each term. Factor it out to get (n-4) * {(n-4) + 1}, so we have (n-4)* (n-3).