Divide using synthetic division $$ ( 5x^{3}-2x^{2}+9x-3 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&5&-2&9&-3\\& & -15& 51& \color{black}{-180} \\ \hline &\color{blue}{5}&\color{blue}{-17}&\color{blue}{60}&\color{orangered}{-183} \end{array} $$The solution is:
$$ \frac{ 5x^{3}-2x^{2}+9x-3 }{ x+3 } = \color{blue}{5x^{2}-17x+60} \color{red}{~-~} \frac{ \color{red}{ 183 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&5&-2&9&-3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 5 }&-2&9&-3\\& & & & \\ \hline &\color{orangered}{5}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&5&-2&9&-3\\& & \color{blue}{-15} & & \\ \hline &\color{blue}{5}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ -17 } $
$$ \begin{array}{c|rrrr}-3&5&\color{orangered}{ -2 }&9&-3\\& & \color{orangered}{-15} & & \\ \hline &5&\color{orangered}{-17}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -17 \right) } = \color{blue}{ 51 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&5&-2&9&-3\\& & -15& \color{blue}{51} & \\ \hline &5&\color{blue}{-17}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ 51 } = \color{orangered}{ 60 } $
$$ \begin{array}{c|rrrr}-3&5&-2&\color{orangered}{ 9 }&-3\\& & -15& \color{orangered}{51} & \\ \hline &5&-17&\color{orangered}{60}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 60 } = \color{blue}{ -180 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&5&-2&9&-3\\& & -15& 51& \color{blue}{-180} \\ \hline &5&-17&\color{blue}{60}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -180 \right) } = \color{orangered}{ -183 } $
$$ \begin{array}{c|rrrr}-3&5&-2&9&\color{orangered}{ -3 }\\& & -15& 51& \color{orangered}{-180} \\ \hline &\color{blue}{5}&\color{blue}{-17}&\color{blue}{60}&\color{orangered}{-183} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{2}-17x+60 } $ with a remainder of $ \color{red}{ -183 } $.