Answer
For the long division problem $2x+1\overset{3x-5}{\overline{\left){\begin{align}
& 6{{x}^{2}}-7x+4 \\
& \underline{6{{x}^{2}}+3x} \\
& \text{ }-10x+4 \\
& \text{ }\underline{-10x-5}\text{ } \\
\end{align}}\right.}}$,
the last term is $9$. Thus, the quotient is $3x-5$ and the remainder is $9$. The answer to this long division problem is $\left( 3x-5 \right)+\frac{9}{2x+1}$.
Work Step by Step
Let us consider,
$2x+1\overset{3x-5}{\overline{\left){\begin{align}
& 6{{x}^{2}}-7x+4 \\
& \underline{6{{x}^{2}}+3x} \\
& \text{ }-10x+4 \\
& \text{ }\underline{-10x-5}\text{ } \\
\end{align}}\right.}}$
Dividend:
$6{{x}^{2}}-7x+4$
Divisor:
$2x+1$
We divide:
$2x+1\overset{3x-5}{\overline{\left){\begin{align}
& 6{{x}^{2}}-7x+4 \\
& \underline{6{{x}^{2}}+3x} \\
& \text{ }-10x+4 \\
& \text{ }\underline{-10x-5} \\
& \text{ }9\text{ } \\
\end{align}}\right.}}$
Thus, the last term is $9$. And the quotient is $3x-5$ and the remainder is $9$. The answer to this long division problem is $\left( 3x-5 \right)+\frac{9}{2x+1}$.
