Find (r-s) (t-s) + (s-r) (s-t) for all numbers r, s, and t. (a) 0 (b) 2 (c) 2rt (d) 2 (s-r) (t-s) (e) 2 (r-s) (t-s)

Notice that the 2 expressions have 2 common terms.

(r-s) is just (s-r) times (-1)

similarly

(t-s) is just (s-t) times (-1)

this means that:

(r-s) (t-s) + (s-r) (s-t) = - (s-r) [ - (s-t) ] + (s-r) (s-t)

the 2 minuses in the first multiplication cancel each other so we have:

- (s-r) [ - (s-t) ] + (s-r) (s-t) = (s-r) (s-t) + (s-r) (s-t) = 2 (s-r) (s-t)

Answer:

d) 2 (s-r) (t-s)