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