Answer
$R=8$
$x-2$ is not factor
Work Step by Step
$$f\left( x\right) =\left( x-c\right) q\left( x\right) +R$$
$$\Rightarrow f\left( c\right) =\left( c-c\right) q\left( x\right) +R$$
$$\Rightarrow R=f\left( c\right) $$
$$f\left( x\right) =4x^{3}-3x^{2}-8x+4=\left( x-2\right) {q}\left( x\right) +R$$
$$\Rightarrow R=f\left( 2\right) =4\times 2^{3}-3\times 2^{2}-8\times 2+4=8\neq 0$$
The Factor Theorem states that if $f(c) = 0,$ then $x - c$ is a factor.
since $f(2)=8\ne0$, $x-2$ is not factor
