Answer:
B only
Step-by-step explanation:
well to solve this question we can consider factor theorem which states that if a Polynomial f(x) is divided by a monomial x-a then x-a is a factor of f(x) if and only if f(a)=0
and to figure "a " we can consider the following equation:
[tex] \displaystyle x - a = 0[/tex]
substitute the given value of a:
[tex] \displaystyle x - 2= 0[/tex]
therefore,
[tex] \displaystyle x = 2[/tex]
option-1:
plugin the value of x to the function A(x):
[tex] \displaystyle A(2) = {2}^{3} + {2}^{2} + 4[/tex]
simplify which yields:
[tex] \displaystyle A(2) = 16[/tex]
since A(2)≠0 x-2 is not a factor of A(x)
option-2:
similarly substitute the value of x to the function B(x) which yields:
[tex] \displaystyle B(2) = {2}^{3} - {2}^{} - 6[/tex]
simplify:
[tex] \displaystyle B(2) =0[/tex]
so, B(2)=0 Hence x-2 is a factor of x³-x-6
option-3:
substitute the value of x to the function C(x)
[tex] \displaystyle C(2) = {2}^{4} + 3.{2}^{} - 10[/tex]
simplify:
[tex] \displaystyle C(2) =12[/tex]
as C(2)≠0 x-2 is not a factor of C(x)
option-4:
likewise plugin the value of x to the function D(x):
[tex] \displaystyle D(2) = {2}^{4} - 2.{2}^{3} - 2[/tex]
simplify and that yields:
[tex] \displaystyle D(2) = - 2[/tex]
therefore,D(2)≠0 hence,x-2 isn't a factor of D(-2)
