Divide using synthetic division $$ ( 4x^{4}-14x^{3}+x^{2}+17x-6 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&4&-14&1&17&-6\\& & 12& -6& -15& \color{black}{6} \\ \hline &\color{blue}{4}&\color{blue}{-2}&\color{blue}{-5}&\color{blue}{2}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 4x^{4}-14x^{3}+x^{2}+17x-6 }{ x-3 } = \color{blue}{4x^{3}-2x^{2}-5x+2} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&-14&1&17&-6\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 4 }&-14&1&17&-6\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 4 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&-14&1&17&-6\\& & \color{blue}{12} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 12 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}3&4&\color{orangered}{ -14 }&1&17&-6\\& & \color{orangered}{12} & & & \\ \hline &4&\color{orangered}{-2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&-14&1&17&-6\\& & 12& \color{blue}{-6} & & \\ \hline &4&\color{blue}{-2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}3&4&-14&\color{orangered}{ 1 }&17&-6\\& & 12& \color{orangered}{-6} & & \\ \hline &4&-2&\color{orangered}{-5}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&-14&1&17&-6\\& & 12& -6& \color{blue}{-15} & \\ \hline &4&-2&\color{blue}{-5}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 17 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}3&4&-14&1&\color{orangered}{ 17 }&-6\\& & 12& -6& \color{orangered}{-15} & \\ \hline &4&-2&-5&\color{orangered}{2}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&-14&1&17&-6\\& & 12& -6& -15& \color{blue}{6} \\ \hline &4&-2&-5&\color{blue}{2}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 6 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}3&4&-14&1&17&\color{orangered}{ -6 }\\& & 12& -6& -15& \color{orangered}{6} \\ \hline &\color{blue}{4}&\color{blue}{-2}&\color{blue}{-5}&\color{blue}{2}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}-2x^{2}-5x+2 } $ with a remainder of $ \color{red}{ 0 } $.