Divide using synthetic division $$ ( 4x^{4}-2x^{2}-8x+3 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&4&0&-2&-8&3\\& & -16& 64& -248& \color{black}{1024} \\ \hline &\color{blue}{4}&\color{blue}{-16}&\color{blue}{62}&\color{blue}{-256}&\color{orangered}{1027} \end{array} $$The solution is:
$$ \frac{ 4x^{4}-2x^{2}-8x+3 }{ x+4 } = \color{blue}{4x^{3}-16x^{2}+62x-256} ~+~ \frac{ \color{red}{ 1027 } }{ x+4 } $$Explanation
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.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&4&0&-2&-8&3\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 4 }&0&-2&-8&3\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 4 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&4&0&-2&-8&3\\& & \color{blue}{-16} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrrr}-4&4&\color{orangered}{ 0 }&-2&-8&3\\& & \color{orangered}{-16} & & & \\ \hline &4&\color{orangered}{-16}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -16 \right) } = \color{blue}{ 64 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&4&0&-2&-8&3\\& & -16& \color{blue}{64} & & \\ \hline &4&\color{blue}{-16}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 64 } = \color{orangered}{ 62 } $
$$ \begin{array}{c|rrrrr}-4&4&0&\color{orangered}{ -2 }&-8&3\\& & -16& \color{orangered}{64} & & \\ \hline &4&-16&\color{orangered}{62}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 62 } = \color{blue}{ -248 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&4&0&-2&-8&3\\& & -16& 64& \color{blue}{-248} & \\ \hline &4&-16&\color{blue}{62}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ \left( -248 \right) } = \color{orangered}{ -256 } $
$$ \begin{array}{c|rrrrr}-4&4&0&-2&\color{orangered}{ -8 }&3\\& & -16& 64& \color{orangered}{-248} & \\ \hline &4&-16&62&\color{orangered}{-256}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -256 \right) } = \color{blue}{ 1024 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&4&0&-2&-8&3\\& & -16& 64& -248& \color{blue}{1024} \\ \hline &4&-16&62&\color{blue}{-256}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 1024 } = \color{orangered}{ 1027 } $
$$ \begin{array}{c|rrrrr}-4&4&0&-2&-8&\color{orangered}{ 3 }\\& & -16& 64& -248& \color{orangered}{1024} \\ \hline &\color{blue}{4}&\color{blue}{-16}&\color{blue}{62}&\color{blue}{-256}&\color{orangered}{1027} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}-16x^{2}+62x-256 } $ with a remainder of $ \color{red}{ 1027 } $.