Question 593861
(3a^2+7a-20)/(a-4):
first factorise 3a^2+7a-20
3a^2+12a-5a-20
(3a^2+12a)-(5a-20)
3a(a+4)-5(a+4)
hence the factors are 
(3a-5)(a+4). THEN divide the factors by (a+4)
={(3a-5)(a+4)}/(a+4)
=(3a-5) because (a+4) will cancel.
.'. our answer is 3a-5.