Divide using synthetic division $$ ( 2x^{4}-10x^{2}+16x+13 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&2&0&-10&16&13\\& & 6& 18& 24& \color{black}{120} \\ \hline &\color{blue}{2}&\color{blue}{6}&\color{blue}{8}&\color{blue}{40}&\color{orangered}{133} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-10x^{2}+16x+13 }{ x-3 } = \color{blue}{2x^{3}+6x^{2}+8x+40} ~+~ \frac{ \color{red}{ 133 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&0&-10&16&13\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 2 }&0&-10&16&13\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&0&-10&16&13\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 6 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}3&2&\color{orangered}{ 0 }&-10&16&13\\& & \color{orangered}{6} & & & \\ \hline &2&\color{orangered}{6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 6 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&0&-10&16&13\\& & 6& \color{blue}{18} & & \\ \hline &2&\color{blue}{6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 18 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}3&2&0&\color{orangered}{ -10 }&16&13\\& & 6& \color{orangered}{18} & & \\ \hline &2&6&\color{orangered}{8}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 8 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&0&-10&16&13\\& & 6& 18& \color{blue}{24} & \\ \hline &2&6&\color{blue}{8}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ 24 } = \color{orangered}{ 40 } $
$$ \begin{array}{c|rrrrr}3&2&0&-10&\color{orangered}{ 16 }&13\\& & 6& 18& \color{orangered}{24} & \\ \hline &2&6&8&\color{orangered}{40}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 40 } = \color{blue}{ 120 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&2&0&-10&16&13\\& & 6& 18& 24& \color{blue}{120} \\ \hline &2&6&8&\color{blue}{40}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 13 } + \color{orangered}{ 120 } = \color{orangered}{ 133 } $
$$ \begin{array}{c|rrrrr}3&2&0&-10&16&\color{orangered}{ 13 }\\& & 6& 18& 24& \color{orangered}{120} \\ \hline &\color{blue}{2}&\color{blue}{6}&\color{blue}{8}&\color{blue}{40}&\color{orangered}{133} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+6x^{2}+8x+40 } $ with a remainder of $ \color{red}{ 133 } $.