Divide using synthetic division $$ ( x^{4}+7x^{3}+13x^{2}+6x-45 ) \div ( x+5 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-5&1&7&13&6&-45\\& & -5& -10& -15& \color{black}{45} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{3}&\color{blue}{-9}&\color{orangered}{0} \end{array} $$The solution is:
$$ \dfrac{ x^{4}+7x^{3}+13x^{2}+6x-45 }{ x+5 } = \color{blue}{x^{3}+2x^{2}+3x-9} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&1&7&13&6&-45\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-5&\color{orangered}{ 1 }&7&13&6&-45\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 1 } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&1&7&13&6&-45\\& & \color{blue}{-5} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}-5&1&\color{orangered}{ 7 }&13&6&-45\\& & \color{orangered}{-5} & & & \\ \hline &1&\color{orangered}{2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 2 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&1&7&13&6&-45\\& & -5& \color{blue}{-10} & & \\ \hline &1&\color{blue}{2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}-5&1&7&\color{orangered}{ 13 }&6&-45\\& & -5& \color{orangered}{-10} & & \\ \hline &1&2&\color{orangered}{3}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 3 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&1&7&13&6&-45\\& & -5& -10& \color{blue}{-15} & \\ \hline &1&2&\color{blue}{3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrrr}-5&1&7&13&\color{orangered}{ 6 }&-45\\& & -5& -10& \color{orangered}{-15} & \\ \hline &1&2&3&\color{orangered}{-9}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ 45 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&1&7&13&6&-45\\& & -5& -10& -15& \color{blue}{45} \\ \hline &1&2&3&\color{blue}{-9}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -45 } + \color{orangered}{ 45 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-5&1&7&13&6&\color{orangered}{ -45 }\\& & -5& -10& -15& \color{orangered}{45} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{3}&\color{blue}{-9}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+2x^{2}+3x-9 } $ with a remainder of $ \color{red}{ 0 } $.