Divide using synthetic division $$ ( -6x^{3}+x^{2}+7x-5 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&-6&1&7&-5\\& & 18& -57& \color{black}{150} \\ \hline &\color{blue}{-6}&\color{blue}{19}&\color{blue}{-50}&\color{orangered}{145} \end{array} $$The solution is:
$$ \frac{ -6x^{3}+x^{2}+7x-5 }{ x+3 } = \color{blue}{-6x^{2}+19x-50} ~+~ \frac{ \color{red}{ 145 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&-6&1&7&-5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ -6 }&1&7&-5\\& & & & \\ \hline &\color{orangered}{-6}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&-6&1&7&-5\\& & \color{blue}{18} & & \\ \hline &\color{blue}{-6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 18 } = \color{orangered}{ 19 } $
$$ \begin{array}{c|rrrr}-3&-6&\color{orangered}{ 1 }&7&-5\\& & \color{orangered}{18} & & \\ \hline &-6&\color{orangered}{19}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 19 } = \color{blue}{ -57 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&-6&1&7&-5\\& & 18& \color{blue}{-57} & \\ \hline &-6&\color{blue}{19}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -57 \right) } = \color{orangered}{ -50 } $
$$ \begin{array}{c|rrrr}-3&-6&1&\color{orangered}{ 7 }&-5\\& & 18& \color{orangered}{-57} & \\ \hline &-6&19&\color{orangered}{-50}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -50 \right) } = \color{blue}{ 150 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&-6&1&7&-5\\& & 18& -57& \color{blue}{150} \\ \hline &-6&19&\color{blue}{-50}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 150 } = \color{orangered}{ 145 } $
$$ \begin{array}{c|rrrr}-3&-6&1&7&\color{orangered}{ -5 }\\& & 18& -57& \color{orangered}{150} \\ \hline &\color{blue}{-6}&\color{blue}{19}&\color{blue}{-50}&\color{orangered}{145} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -6x^{2}+19x-50 } $ with a remainder of $ \color{red}{ 145 } $.