Divide using synthetic division $$ ( 3x^{4}-22x^{3}+13x^{2}+118x-40 ) \div ( x-5 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&3&-22&13&118&-40\\& & 15& -35& -110& \color{black}{40} \\ \hline &\color{blue}{3}&\color{blue}{-7}&\color{blue}{-22}&\color{blue}{8}&\color{orangered}{0} \end{array} $$The solution is:
$$ \dfrac{ 3x^{4}-22x^{3}+13x^{2}+118x-40 }{ x-5 } = \color{blue}{3x^{3}-7x^{2}-22x+8} $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&-22&13&118&-40\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ 3 }&-22&13&118&-40\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 3 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&-22&13&118&-40\\& & \color{blue}{15} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -22 } + \color{orangered}{ 15 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrr}5&3&\color{orangered}{ -22 }&13&118&-40\\& & \color{orangered}{15} & & & \\ \hline &3&\color{orangered}{-7}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -35 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&-22&13&118&-40\\& & 15& \color{blue}{-35} & & \\ \hline &3&\color{blue}{-7}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ \left( -35 \right) } = \color{orangered}{ -22 } $
$$ \begin{array}{c|rrrrr}5&3&-22&\color{orangered}{ 13 }&118&-40\\& & 15& \color{orangered}{-35} & & \\ \hline &3&-7&\color{orangered}{-22}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -22 \right) } = \color{blue}{ -110 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&-22&13&118&-40\\& & 15& -35& \color{blue}{-110} & \\ \hline &3&-7&\color{blue}{-22}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 118 } + \color{orangered}{ \left( -110 \right) } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}5&3&-22&13&\color{orangered}{ 118 }&-40\\& & 15& -35& \color{orangered}{-110} & \\ \hline &3&-7&-22&\color{orangered}{8}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 8 } = \color{blue}{ 40 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&3&-22&13&118&-40\\& & 15& -35& -110& \color{blue}{40} \\ \hline &3&-7&-22&\color{blue}{8}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -40 } + \color{orangered}{ 40 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}5&3&-22&13&118&\color{orangered}{ -40 }\\& & 15& -35& -110& \color{orangered}{40} \\ \hline &\color{blue}{3}&\color{blue}{-7}&\color{blue}{-22}&\color{blue}{8}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}-7x^{2}-22x+8 } $ with a remainder of $ \color{red}{ 0 } $.