Question

See screenshot below​

Answer 1

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:

$\displaystyle x - a = 0$

substitute the given value of a:

$\displaystyle x - 2= 0$

therefore,

$\displaystyle x = 2$

option-1:

plugin the value of x to the function A(x):

$\displaystyle A(2) = {2}^{3} + {2}^{2} + 4$

simplify which yields:

$\displaystyle A(2) = 16$

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:

$\displaystyle B(2) = {2}^{3} - {2}^{} - 6$

simplify:

$\displaystyle B(2) =0$

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)

$\displaystyle C(2) = {2}^{4} + 3.{2}^{} - 10$

simplify:

$\displaystyle C(2) =12$

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):

$\displaystyle D(2) = {2}^{4} - 2.{2}^{3} - 2$

simplify and that yields:

$\displaystyle D(2) = - 2$

therefore,D(2)≠0 hence,x-2 isn't a factor of D(-2)

Answer 2

$\boxed{\large{\bold{\textbf{\textsf{{\color{blue}{Answer}}}}}}:)}$

The factor

• x-2=0
• x=2

we have find the value of x .

Now,

we have to select all the polynomial which has x-2 As a factor.

According to the question,

option A:-

$A(x)=x^3+x^2+4\\\\A(2)=(2)^3+(2)^2+4\\\\=8+4+4\\\\16≠0$

• so it is not a factor.

Option B:-

$B(x)=x^3-x-6\\\\=B(2)=(2)^2-2-6\\\\0=0$

• so it is a factor

Option C:-

$C(x)=x^4+3x-10\\\\=C(2)=(2)^4+3(2)-10\\\\=16+6-10\\\\12≠0$

• so it is not a factor

option D:-

$D(x)=x^4-2x^3-2\\\\D(2)=(2)^4-2(2)^3-2\\\\=16-16-2\\\\-2≠0$

• so it is not a factor

therefore:-

The option B is correct.