Respuesta :
The remainder theorem says that if [tex]x-c[/tex] is a factor of a polynomial [tex]p(x)[/tex], then the remainder upon dividing [tex]\dfrac{p(x)}{x-c}[/tex] is 0 so that there exists a lower degree polynomial [tex]q(x)[/tex] as the quotient:
[tex]\dfrac{p(x)}{x-c}=q(x)[/tex]
Using the fact that [tex]x-2[/tex] is a factor, you can find a quadratic [tex]q(x)[/tex] which is easy to factorize further.
Synthetic division yields
[tex]q(x)=4x^2+4x-8[/tex]
which can be factored further as
[tex]4(x^2+x-2)=4(x+2)(x-1)[/tex]
So,
[tex]f(x)=4x^3-4x^2-16x+16=4(x-2)(x+2)(x-1)[/tex]
The roots are then [tex]x=-2,1,2[/tex].
[tex]\dfrac{p(x)}{x-c}=q(x)[/tex]
Using the fact that [tex]x-2[/tex] is a factor, you can find a quadratic [tex]q(x)[/tex] which is easy to factorize further.
Synthetic division yields
[tex]q(x)=4x^2+4x-8[/tex]
which can be factored further as
[tex]4(x^2+x-2)=4(x+2)(x-1)[/tex]
So,
[tex]f(x)=4x^3-4x^2-16x+16=4(x-2)(x+2)(x-1)[/tex]
The roots are then [tex]x=-2,1,2[/tex].
Answer:
The answer to this question is "B.) x = -2, x = 1, or x = 2"