Divide using synthetic division $$ ( 5x^{4}-x^{3}+x^{2}-5 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&5&-1&1&0&-5\\& & -15& 48& -147& \color{black}{441} \\ \hline &\color{blue}{5}&\color{blue}{-16}&\color{blue}{49}&\color{blue}{-147}&\color{orangered}{436} \end{array} $$The solution is:
$$ \frac{ 5x^{4}-x^{3}+x^{2}-5 }{ x+3 } = \color{blue}{5x^{3}-16x^{2}+49x-147} ~+~ \frac{ \color{red}{ 436 } }{ x+3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&5&-1&1&0&-5\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 5 }&-1&1&0&-5\\& & & & & \\ \hline &\color{orangered}{5}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 5 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&5&-1&1&0&-5\\& & \color{blue}{-15} & & & \\ \hline &\color{blue}{5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrrr}-3&5&\color{orangered}{ -1 }&1&0&-5\\& & \color{orangered}{-15} & & & \\ \hline &5&\color{orangered}{-16}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -16 \right) } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&5&-1&1&0&-5\\& & -15& \color{blue}{48} & & \\ \hline &5&\color{blue}{-16}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 48 } = \color{orangered}{ 49 } $
$$ \begin{array}{c|rrrrr}-3&5&-1&\color{orangered}{ 1 }&0&-5\\& & -15& \color{orangered}{48} & & \\ \hline &5&-16&\color{orangered}{49}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 49 } = \color{blue}{ -147 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&5&-1&1&0&-5\\& & -15& 48& \color{blue}{-147} & \\ \hline &5&-16&\color{blue}{49}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -147 \right) } = \color{orangered}{ -147 } $
$$ \begin{array}{c|rrrrr}-3&5&-1&1&\color{orangered}{ 0 }&-5\\& & -15& 48& \color{orangered}{-147} & \\ \hline &5&-16&49&\color{orangered}{-147}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -147 \right) } = \color{blue}{ 441 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&5&-1&1&0&-5\\& & -15& 48& -147& \color{blue}{441} \\ \hline &5&-16&49&\color{blue}{-147}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 441 } = \color{orangered}{ 436 } $
$$ \begin{array}{c|rrrrr}-3&5&-1&1&0&\color{orangered}{ -5 }\\& & -15& 48& -147& \color{orangered}{441} \\ \hline &\color{blue}{5}&\color{blue}{-16}&\color{blue}{49}&\color{blue}{-147}&\color{orangered}{436} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{3}-16x^{2}+49x-147 } $ with a remainder of $ \color{red}{ 436 } $.