Divide using synthetic division $$ ( 3x^{3}-17x^{2}+13x+23 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&3&-17&13&23\\& & 12& -20& \color{black}{-28} \\ \hline &\color{blue}{3}&\color{blue}{-5}&\color{blue}{-7}&\color{orangered}{-5} \end{array} $$The solution is:
$$ \frac{ 3x^{3}-17x^{2}+13x+23 }{ x-4 } = \color{blue}{3x^{2}-5x-7} \color{red}{~-~} \frac{ \color{red}{ 5 } }{ x-4 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{4}&3&-17&13&23\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 3 }&-17&13&23\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 3 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&3&-17&13&23\\& & \color{blue}{12} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -17 } + \color{orangered}{ 12 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}4&3&\color{orangered}{ -17 }&13&23\\& & \color{orangered}{12} & & \\ \hline &3&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&3&-17&13&23\\& & 12& \color{blue}{-20} & \\ \hline &3&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}4&3&-17&\color{orangered}{ 13 }&23\\& & 12& \color{orangered}{-20} & \\ \hline &3&-5&\color{orangered}{-7}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ -28 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&3&-17&13&23\\& & 12& -20& \color{blue}{-28} \\ \hline &3&-5&\color{blue}{-7}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 23 } + \color{orangered}{ \left( -28 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}4&3&-17&13&\color{orangered}{ 23 }\\& & 12& -20& \color{orangered}{-28} \\ \hline &\color{blue}{3}&\color{blue}{-5}&\color{blue}{-7}&\color{orangered}{-5} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}-5x-7 } $ with a remainder of $ \color{red}{ -5 } $.