Synthetic Division - Regular Synthetic Division

Regular Synthetic Division

The first example is synthetic division with only a monic linear denominator .

Write the coefficients of the polynomial to be divided at the top (the zero is for the unseen 0x).

$\begin{array}{cc} \begin{array}{r} \\ \\ \end{array} & \begin{array}{|rrrr} 1 & -12 & 0 & -42 \\ & & & \\ \hline \end{array} \end{array}$

Negate the coefficients of the divisor.

$\begin{array}{rr} -1x & + 3 \end{array}$

Write in every coefficient of the divisor but the first one on the left.

$\begin{array}{cc} \begin{array}{r} \\ 3 \\ \end{array} & \begin{array}{|rrrr} 1 & -12 & 0 & -42 \\ & & & \\ \hline \end{array} \end{array}$

Note the change of sign from −3 to 3. "Drop" the first coefficient after the bar to the last row.

$\begin{array}{cc} \begin{array}{r} \\ 3 \\ \\ \end{array} & \begin{array}{|rrrr} 1 & -12 & 0 & -42 \\ & & & \\ \hline 1 & & & \\ \end{array} \end{array}$

Multiply the dropped number by the number before the bar, and place it in the next column.

$\begin{array}{cc} \begin{array}{r} \\ 3 \\ \\ \end{array} & \begin{array}{|rrrr} 1 & -12 & 0 & -42 \\ & 3 & & \\ \hline 1 & & & \\ \end{array} \end{array}$

Perform an addition in the next column.

$\begin{array}{cc} \begin{array}{c} \\ 3 \\ \\ \end{array} & \begin{array}{|rrrr} 1 & -12 & 0 & -42 \\ & 3 & & \\ \hline 1 & -9 & & \\ \end{array} \end{array}$

Repeat the previous two steps and the following is obtained:

$\begin{array}{cc} \begin{array}{c} \\ 3 \\ \\ \end{array} & \begin{array}{|rrrr} 1 & -12 & 0 & -42 \\ & 3 & -27 & -81 \\ \hline 1 & -9 & -27 & -123 \end{array} \end{array}$

Count the terms to the left of the bar. Since there is only one, the remainder has degree zero. Mark the separation with a vertical bar.

$\begin{array}{rrr|r} 1 & -9 & -27 & -123 \end{array}$

The terms are written with increasing degree from right to left beginning with degree zero for both the remainder and the result.

$\begin{array}{rrr|r} 1x^2 & -9x & -27 & -123 \end{array}$

The result of our division is:

Evaluating Polynomials by the Remainder Theorem

The above form of synthetic division is useful in the context of the Polynomial remainder theorem for evaluating univariate polynomials. To summarize, the value of at is equal to the remainder of . The advantage of calculating the value this way is that it requires just over half as many multiplication steps as naive evaluation. An alternative evaluation strategy is Horner's method.

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