Respuesta :
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)
(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)