How do you use the factor theorem to determine whether $x-2$ is a factor of \[4{{x}^{3}}-3{{x}^{2}}-8x+4\]?

Hint: In this question we have been given with a polynomial expression and we have to show using the factor theorem that $x-2$ is a factor of the given polynomial equation. We will first find the root of the equation using the factor of the equation. And then we will use the factor theorem which states that $\left( x-a \right)$ is a factor of $f\left( x \right)$ given $f\left( a \right)=0$. We will substitute the value of the root in the polynomial, if the value is $0$, $x-2$ is a factor.

Complete step-by-step answer:
We have the expression given to us as:
\[\Rightarrow 4{{x}^{3}}-3{{x}^{2}}-8x+4\]
We have to show that $x-2$ is a factor of the equation using the factor theorem.
The factor theorem states that $\left( x-a \right)$ is a factor of $f\left( x \right)$ if $f\left( a \right)=0$.
We have the term $\left( x-a \right)$ given to us as:
$\Rightarrow x-2$
therefore, the value of $a$ will be:
$\Rightarrow a=2$
Now the function is given to us as:
\[\Rightarrow f\left( x \right)=4{{x}^{3}}-3{{x}^{2}}-8x+4\]
On substituting the value of $x=2$, we get:
$\Rightarrow f\left( 2 \right)=4{{\left( 2 \right)}^{3}}-3{{\left( 2 \right)}^{2}}-8\left( 2 \right)+4$
On simplifying the brackets, we get:
$\Rightarrow f\left( 2 \right)=4\times 8-3\times 4-16+4$
On multiplying the terms, we get:
$\Rightarrow f\left( 2 \right)=32-12-16+4$
On simplifying, we get:
$\Rightarrow f\left( 2 \right)=8$
Since $f\left( a \right)\ne 0$, we can conclude that $x-2$ is not a factor of the given polynomial, which is the required solution.

Note: It is to be remembered that the factor theorem is a special form of the remainder theorem. Since the value of the function is $8$ it is the remainder of the given polynomial. Since the polynomial has the highest exponent as $3$, the degree of the given polynomial is $3$. It is to be remembered that a factor is a term which divides a given number without leaving any remainder.