How do you define a polynomial #p(x)# that is zero or negative whenever #x in [-2, 1] uu [4, oo)# ?

1 Answer
Mar 12, 2017

Answer:

#p(x) = -(x+2)(x-1)(x-4) = -x^3+3x^2+6x-8#

Explanation:

Define:

#p(x) = -(x+2)(x-1)(x-4) = -x^3+3x^2+6x-8#

graph{ -x^3+3x^2+6x-8 [-10, 10, -15, 15]}

This cubic has a leading negative coefficient #-1# and zeros at #x=-2#, #x=1# and #x=4#.

As a result, it satisfies the required conditions.

Any polynomial in #x# with these zeros will be a multiple (scalar or polynomial) of #p(x)#.

Any positive scalar multiple of #p(x)# will also be a cubic satisfying the conditions.

If #p(x)# is multiplied by any polynomial #q(x)# satisfying the following conditions, then the resulting polynomial also satisfies the conditions:

#{ (q(x) > 0 " for all " x in (-oo, -2) uu (-2, 1) uu (1, 4) uu (4, oo)), (q(x) >= 0 " for all " x in { -2, 1, 4 }) :}#

For example, we can multiply #p(x)# by #q(x) = (x-1)^2(x^2+2)# to get a septic polynomial that is non-positive in #[-2, 1] uu [4, oo)#.

graph{(-x^3+3x^2+6x-8)(x-1)^2(x^2+2) [-10, 10, -240, 700]}