The synthetic division table is:
$$ \begin{array}{c|rrrr}0&3&7&-116&160\\& & 0& 0& \color{black}{0} \\ \hline &\color{blue}{3}&\color{blue}{7}&\color{blue}{-116}&\color{orangered}{160} \end{array} $$The remainder when $ 3x^{3}+7x^{2}-116x+160 $ is divided by $ x $ is $ \, \color{red}{ 160 } $.
We can find remainder using synthetic division method.
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrr}\color{blue}{0}&3&7&-116&160\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}0&\color{orangered}{ 3 }&7&-116&160\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 3 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&3&7&-116&160\\& & \color{blue}{0} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 0 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}0&3&\color{orangered}{ 7 }&-116&160\\& & \color{orangered}{0} & & \\ \hline &3&\color{orangered}{7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 7 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&3&7&-116&160\\& & 0& \color{blue}{0} & \\ \hline &3&\color{blue}{7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -116 } + \color{orangered}{ 0 } = \color{orangered}{ -116 } $
$$ \begin{array}{c|rrrr}0&3&7&\color{orangered}{ -116 }&160\\& & 0& \color{orangered}{0} & \\ \hline &3&7&\color{orangered}{-116}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -116 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&3&7&-116&160\\& & 0& 0& \color{blue}{0} \\ \hline &3&7&\color{blue}{-116}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 160 } + \color{orangered}{ 0 } = \color{orangered}{ 160 } $
$$ \begin{array}{c|rrrr}0&3&7&-116&\color{orangered}{ 160 }\\& & 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{3}&\color{blue}{7}&\color{blue}{-116}&\color{orangered}{160} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 160 }\right) $.