Divide using synthetic division $$ ( 5x^{3}-2x^{2}+1 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&5&-2&0&1\\& & 15& 39& \color{black}{117} \\ \hline &\color{blue}{5}&\color{blue}{13}&\color{blue}{39}&\color{orangered}{118} \end{array} $$The solution is:
$$ \frac{ 5x^{3}-2x^{2}+1 }{ x-3 } = \color{blue}{5x^{2}+13x+39} ~+~ \frac{ \color{red}{ 118 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&5&-2&0&1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 5 }&-2&0&1\\& & & & \\ \hline &\color{orangered}{5}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 5 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&5&-2&0&1\\& & \color{blue}{15} & & \\ \hline &\color{blue}{5}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 15 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrr}3&5&\color{orangered}{ -2 }&0&1\\& & \color{orangered}{15} & & \\ \hline &5&\color{orangered}{13}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 13 } = \color{blue}{ 39 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&5&-2&0&1\\& & 15& \color{blue}{39} & \\ \hline &5&\color{blue}{13}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 39 } = \color{orangered}{ 39 } $
$$ \begin{array}{c|rrrr}3&5&-2&\color{orangered}{ 0 }&1\\& & 15& \color{orangered}{39} & \\ \hline &5&13&\color{orangered}{39}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 39 } = \color{blue}{ 117 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&5&-2&0&1\\& & 15& 39& \color{blue}{117} \\ \hline &5&13&\color{blue}{39}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 117 } = \color{orangered}{ 118 } $
$$ \begin{array}{c|rrrr}3&5&-2&0&\color{orangered}{ 1 }\\& & 15& 39& \color{orangered}{117} \\ \hline &\color{blue}{5}&\color{blue}{13}&\color{blue}{39}&\color{orangered}{118} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{2}+13x+39 } $ with a remainder of $ \color{red}{ 118 } $.