Is the value of (x-1)(x-2)(x-3)(x-4)+5(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(x−1)(x−2)(x−3)(x−4)+5
= (t+3/2)(t+1/2)(t-1/2)(t-3/2)+5=(t+32)(t+12)(t−12)(t−32)+5
= (t^2-9/4)(t^2-1/4)+5=(t2−94)(t2−14)+5
= t^4-5/2t^2+89/16=t4−52t2+8916
= (t^2)^2-5/2(t^2)+89/16=(t2)2−52(t2)+8916
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]}
