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 #x#

Explanation:

Using the symmetry around #x=5/2#, let #x = t+5/2#

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 #t^2# let us look at its discriminant:

#Delta = b^2-4ac = (-5/2)^2-4(1)(89/16) = 25/4-89/4 = -16#

Since #Delta < 0#, this quadratic has no Real solutions.

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]}