Divide using synthetic division $$ ( 14x^{3}+x^{2}-30x+13 ) \div ( 7x-10 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrr}10&14&1&-30&13\\& & 140& 1410& \color{black}{13800} \\ \hline &\color{blue}{14}&\color{blue}{141}&\color{blue}{1380}&\color{orangered}{13813} \end{array} $$The solution is:
$$ \dfrac{ 14x^{3}+x^{2}-30x+13 }{ x-10 } = \color{blue}{14x^{2}+141x+1380} ~+~ \dfrac{ \color{red}{ 13813 } }{ x-10 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -10 = 0 $ ( $ x = \color{blue}{ 10 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{10}&14&1&-30&13\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}10&\color{orangered}{ 14 }&1&-30&13\\& & & & \\ \hline &\color{orangered}{14}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 10 } \cdot \color{blue}{ 14 } = \color{blue}{ 140 } $.
$$ \begin{array}{c|rrrr}\color{blue}{10}&14&1&-30&13\\& & \color{blue}{140} & & \\ \hline &\color{blue}{14}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 140 } = \color{orangered}{ 141 } $
$$ \begin{array}{c|rrrr}10&14&\color{orangered}{ 1 }&-30&13\\& & \color{orangered}{140} & & \\ \hline &14&\color{orangered}{141}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 10 } \cdot \color{blue}{ 141 } = \color{blue}{ 1410 } $.
$$ \begin{array}{c|rrrr}\color{blue}{10}&14&1&-30&13\\& & 140& \color{blue}{1410} & \\ \hline &14&\color{blue}{141}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -30 } + \color{orangered}{ 1410 } = \color{orangered}{ 1380 } $
$$ \begin{array}{c|rrrr}10&14&1&\color{orangered}{ -30 }&13\\& & 140& \color{orangered}{1410} & \\ \hline &14&141&\color{orangered}{1380}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 10 } \cdot \color{blue}{ 1380 } = \color{blue}{ 13800 } $.
$$ \begin{array}{c|rrrr}\color{blue}{10}&14&1&-30&13\\& & 140& 1410& \color{blue}{13800} \\ \hline &14&141&\color{blue}{1380}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ 13800 } = \color{orangered}{ 13813 } $
$$ \begin{array}{c|rrrr}10&14&1&-30&\color{orangered}{ 13 }\\& & 140& 1410& \color{orangered}{13800} \\ \hline &\color{blue}{14}&\color{blue}{141}&\color{blue}{1380}&\color{orangered}{13813} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 14x^{2}+141x+1380 } $ with a remainder of $ \color{red}{ 13813 } $.