#### Answer

The given statement is true.

#### Work Step by Step

Let us try to perform synthetic division over division of $f\left( x \right)$ by $g\left( x \right)$.
$\frac{1}{2}\overline{\left){\begin{align}
& 10\,\,\,\,\,-6\,\,\,\,\,\,\,4\,\,\,\,\,\,-1 \\
& \frac{\downarrow \,\,\,\,\,\,\,\,\,5\,\,\,\,\,-\frac{1}{2}\,\,\,\,\,\,\,\frac{7}{4}\,}{\begin{align}
& 10\,\,\,\,\,-1\,\,\,\,\,\,\,\,\frac{7}{2}\,\,\,\,\,\,\,\,\frac{3}{4}\, \\
& \frac{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow \,\,}{10{{x}^{2}}-x+\frac{7}{2}\,\,\,\,\,\,\,R}\, \\
\end{align}} \\
\end{align}}\right.}$
Here, the quotient obtained is $h\left( x \right)=10{{x}^{2}}-x+\frac{7}{2}$ and the remainder is $\frac{3}{4}$.
By checking out the solution:
$\begin{align}
& f\left( x \right)=g\left( x \right).h\left( x \right)+R \\
& =\left( x-\frac{1}{2} \right)\left( 10{{x}^{2}}-x+\frac{7}{2} \right)+\frac{3}{4} \\
& =10{{x}^{3}}-{{x}^{2}}+\frac{7}{2}x-5{{x}^{2}}+\frac{1}{2}x-\frac{7}{4}+\frac{3}{4} \\
& =10{{x}^{3}}-6{{x}^{2}}+4x-1.
\end{align}$
Therefore, synthetic division can be used to find the quotient of $f\left( x \right)=10{{x}^{3}}-6{{x}^{2}}+4x-1$ when divided by $g\left( x \right)=x-\frac{1}{2}$.