Divide using synthetic division $$ ( 2x^{3}-13x^{2}+10 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&2&-13&0&10\\& & 10& -15& \color{black}{-75} \\ \hline &\color{blue}{2}&\color{blue}{-3}&\color{blue}{-15}&\color{orangered}{-65} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-13x^{2}+10 }{ x-5 } = \color{blue}{2x^{2}-3x-15} \color{red}{~-~} \frac{ \color{red}{ 65 } }{ x-5 } $$Explanation
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.
$$ \begin{array}{c|rrrr}\color{blue}{5}&2&-13&0&10\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 2 }&-13&0&10\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 2 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&2&-13&0&10\\& & \color{blue}{10} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 10 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}5&2&\color{orangered}{ -13 }&0&10\\& & \color{orangered}{10} & & \\ \hline &2&\color{orangered}{-3}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&2&-13&0&10\\& & 10& \color{blue}{-15} & \\ \hline &2&\color{blue}{-3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ -15 } $
$$ \begin{array}{c|rrrr}5&2&-13&\color{orangered}{ 0 }&10\\& & 10& \color{orangered}{-15} & \\ \hline &2&-3&\color{orangered}{-15}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -15 \right) } = \color{blue}{ -75 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&2&-13&0&10\\& & 10& -15& \color{blue}{-75} \\ \hline &2&-3&\color{blue}{-15}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -75 \right) } = \color{orangered}{ -65 } $
$$ \begin{array}{c|rrrr}5&2&-13&0&\color{orangered}{ 10 }\\& & 10& -15& \color{orangered}{-75} \\ \hline &\color{blue}{2}&\color{blue}{-3}&\color{blue}{-15}&\color{orangered}{-65} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-3x-15 } $ with a remainder of $ \color{red}{ -65 } $.