Divide using synthetic division $$ ( 4x^{3}-10x^{2}+5 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&4&-10&0&5\\& & 16& 24& \color{black}{96} \\ \hline &\color{blue}{4}&\color{blue}{6}&\color{blue}{24}&\color{orangered}{101} \end{array} $$The solution is:
$$ \frac{ 4x^{3}-10x^{2}+5 }{ x-4 } = \color{blue}{4x^{2}+6x+24} ~+~ \frac{ \color{red}{ 101 } }{ x-4 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{4}&4&-10&0&5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 4 }&-10&0&5\\& & & & \\ \hline &\color{orangered}{4}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 4 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&4&-10&0&5\\& & \color{blue}{16} & & \\ \hline &\color{blue}{4}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 16 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}4&4&\color{orangered}{ -10 }&0&5\\& & \color{orangered}{16} & & \\ \hline &4&\color{orangered}{6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 6 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&4&-10&0&5\\& & 16& \color{blue}{24} & \\ \hline &4&\color{blue}{6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 24 } = \color{orangered}{ 24 } $
$$ \begin{array}{c|rrrr}4&4&-10&\color{orangered}{ 0 }&5\\& & 16& \color{orangered}{24} & \\ \hline &4&6&\color{orangered}{24}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 24 } = \color{blue}{ 96 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&4&-10&0&5\\& & 16& 24& \color{blue}{96} \\ \hline &4&6&\color{blue}{24}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 96 } = \color{orangered}{ 101 } $
$$ \begin{array}{c|rrrr}4&4&-10&0&\color{orangered}{ 5 }\\& & 16& 24& \color{orangered}{96} \\ \hline &\color{blue}{4}&\color{blue}{6}&\color{blue}{24}&\color{orangered}{101} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{2}+6x+24 } $ with a remainder of $ \color{red}{ 101 } $.