Answer: A,B and C are true.
Step-by-step explanation:
let f(x) be the given polynomial with variable x such that
[tex]f(x)=ax(x-2)^4[/tex]m,where a be any odd degree negative leading coefficient of f(x),x has root as 0 with multiplicity 1 and [tex](x-2)^4[/tex] has root 2 with multiplicity 4.
Lets check all the options
A. The function is positive on (-∞, 0)
let x=-1∈(-∞, 0)
⇒[tex]f(x)=a(-1)(-1-2)^4=-a(-3)^4=-81a[/tex]> 0 as a is negative.
∴ function is positive on (-∞, 0) .i,e. A is true.
B. The function is negative on (0, 2).
Let x=1 ∈(0,2)
⇒[tex]f(x)=a(1)(1-2)^4=a(-1)^4=a[/tex]< 0 as a is negative.
∴ the function is negative on (0, 2) .i,e. B is true.
C. The function is negative on (2, ∞)
let x=3∈(2,∞)
⇒[tex]f(x)=a(3)(3-2)^4=3a(1)^4=3a[/tex]< 0 as a is negative.
∴ the function is negative on (2,∞).
D.The function is positive on (0, ∞) which is not true from C.
