Is the value of #(x-1)(x-2)(x-3)(x-4)+5#, positive, negative or zero? Thank you!
1 Answer
Nov 24, 2016
This quartic is always positive for Real values of
Explanation:
Using the symmetry around
Then:
#(x-1)(x-2)(x-3)(x-4)+5#
#= (t+3/2)(t+1/2)(t-1/2)(t-3/2)+5#
#= (t^2-9/4)(t^2-1/4)+5#
#= t^4-5/2t^2+89/16#
#= (t^2)^2-5/2(t^2)+89/16#
Treating this as a quadratic in
#Delta = b^2-4ac = (-5/2)^2-4(1)(89/16) = 25/4-89/4 = -16#
Since
So there are no Real zeros and our original quartic is always positive (since its leading coefficient is positive).
graph{(x-1)(x-2)(x-3)(x-4)+5 [-1, 5.2, -1, 9.16]}