Find all real values of $p$ such that
(x+1)(x-2p)
has a minimum value of 0 over all real values of $x$.
(x + 1) (x-2p) =
x^2 + (1 -2p)x -2p
This is a parabola that turns upward
The x value of the vertex = - (1 -2p) / (2) = -1/2 + p
We want this x value to produce 0...so....
(x + 1) ( x -2p) = 0
( -1/2 + p + 1 ) ( -1/2 + p -2p) = 0
(1/2 + p) ( -1/2 -p) = 0
(1/2)+ p = 0 -1/2 - p = 0
p = -1/2 p =-1/2
p = -1/2