Divide using synthetic division $$ ( 3x^{4}-14x^{3}+x^{2}+26x+8 ) \div ( x-4 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}4&3&-14&1&26&8\\& & 12& -8& -28& \color{black}{-8} \\ \hline &\color{blue}{3}&\color{blue}{-2}&\color{blue}{-7}&\color{blue}{-2}&\color{orangered}{0} \end{array} $$The solution is:
$$ \dfrac{ 3x^{4}-14x^{3}+x^{2}+26x+8 }{ x-4 } = \color{blue}{3x^{3}-2x^{2}-7x-2} $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&3&-14&1&26&8\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}4&\color{orangered}{ 3 }&-14&1&26&8\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 3 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&3&-14&1&26&8\\& & \color{blue}{12} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 12 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}4&3&\color{orangered}{ -14 }&1&26&8\\& & \color{orangered}{12} & & & \\ \hline &3&\color{orangered}{-2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&3&-14&1&26&8\\& & 12& \color{blue}{-8} & & \\ \hline &3&\color{blue}{-2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrr}4&3&-14&\color{orangered}{ 1 }&26&8\\& & 12& \color{orangered}{-8} & & \\ \hline &3&-2&\color{orangered}{-7}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -28 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&3&-14&1&26&8\\& & 12& -8& \color{blue}{-28} & \\ \hline &3&-2&\color{blue}{-7}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 26 } + \color{orangered}{ \left( -28 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}4&3&-14&1&\color{orangered}{ 26 }&8\\& & 12& -8& \color{orangered}{-28} & \\ \hline &3&-2&-7&\color{orangered}{-2}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&3&-14&1&26&8\\& & 12& -8& -28& \color{blue}{-8} \\ \hline &3&-2&-7&\color{blue}{-2}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}4&3&-14&1&26&\color{orangered}{ 8 }\\& & 12& -8& -28& \color{orangered}{-8} \\ \hline &\color{blue}{3}&\color{blue}{-2}&\color{blue}{-7}&\color{blue}{-2}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}-2x^{2}-7x-2 } $ with a remainder of $ \color{red}{ 0 } $.