find the value of n such that the expression is a perfect square trinomial. y^2-40y+n
n=400
Factor the following: y^2-40 y+400
The factors of 400 that sum to -40 are -20 and -20. So, y^2-40 y+400 = (y-20) (y-20): (y-20) (y-20)
(y-20) (y-20) = (y-20)^2: Answer: | (y-20)^2
thx
+7 
