Answer
When $g\left( x \right)=\left( x-2 \right)$ divides $f\left( x \right)=2{{x}^{2}}-7x+9$, the quotient is $2x-3$ and it leaves the remainder $3$.
Work Step by Step
We put in the value to determine the value of the divisor $g\left( x \right)$:
$\begin{align}
& 2{{x}^{2}}-7x+9=g\left( x \right).\left( 2x-3 \right)+3 \\
& g\left( x \right).\left( 2x-3 \right)=2{{x}^{2}}-7x+6 \\
& g\left( x \right)=\frac{2{{x}^{2}}-7x+6}{2x-3} \\
& g\left( x \right)=\frac{\left( 2x-3 \right)\left( x-2 \right)}{\left( 2x-3 \right)} \\
\end{align}$
So,
$g\left( x \right)=\left( x-2 \right)$
Therefore, the divisor is $g\left( x \right)=x-2$.
