Divide using synthetic division $$ ( 15x^{4}+x^{3}-52x^{2}+20x+16 ) \div ( x+2 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&15&1&-52&20&16\\& & -30& 58& -12& \color{black}{-16} \\ \hline &\color{blue}{15}&\color{blue}{-29}&\color{blue}{6}&\color{blue}{8}&\color{orangered}{0} \end{array} $$The solution is:
$$ \dfrac{ 15x^{4}+x^{3}-52x^{2}+20x+16 }{ x+2 } = \color{blue}{15x^{3}-29x^{2}+6x+8} $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&15&1&-52&20&16\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 15 }&1&-52&20&16\\& & & & & \\ \hline &\color{orangered}{15}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 15 } = \color{blue}{ -30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&15&1&-52&20&16\\& & \color{blue}{-30} & & & \\ \hline &\color{blue}{15}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ -29 } $
$$ \begin{array}{c|rrrrr}-2&15&\color{orangered}{ 1 }&-52&20&16\\& & \color{orangered}{-30} & & & \\ \hline &15&\color{orangered}{-29}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -29 \right) } = \color{blue}{ 58 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&15&1&-52&20&16\\& & -30& \color{blue}{58} & & \\ \hline &15&\color{blue}{-29}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -52 } + \color{orangered}{ 58 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}-2&15&1&\color{orangered}{ -52 }&20&16\\& & -30& \color{orangered}{58} & & \\ \hline &15&-29&\color{orangered}{6}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 6 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&15&1&-52&20&16\\& & -30& 58& \color{blue}{-12} & \\ \hline &15&-29&\color{blue}{6}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}-2&15&1&-52&\color{orangered}{ 20 }&16\\& & -30& 58& \color{orangered}{-12} & \\ \hline &15&-29&6&\color{orangered}{8}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 8 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&15&1&-52&20&16\\& & -30& 58& -12& \color{blue}{-16} \\ \hline &15&-29&6&\color{blue}{8}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-2&15&1&-52&20&\color{orangered}{ 16 }\\& & -30& 58& -12& \color{orangered}{-16} \\ \hline &\color{blue}{15}&\color{blue}{-29}&\color{blue}{6}&\color{blue}{8}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 15x^{3}-29x^{2}+6x+8 } $ with a remainder of $ \color{red}{ 0 } $.