Question

Is the value of `(x-1)(x-2)(x-3)(x-4)+5`, positive, negative or zero? Thank you!

Answer 1

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]}