In order to solve this problem, we need to perform synthetic division. We set up synthetic division by writing down the zero of the expression we are dividing by, and the coefficients of the polynomial on a line.

\(\displaystyle \begin{tabular}{ c| ccc}-13&11&-9&7\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}\)

The first step is to bring the coefficient of the \(\displaystyle \uptext{x}^2\) term down.

\(\displaystyle \begin{tabular}{ c| ccc}-13&11&-9&7\\ ~ & ~ & ~ & ~ \\ \hline ~ &11& ~ & ~ \end{tabular}\)

Now we multiply the zero by the term we just put down, and place it under the \(\displaystyle \uptext{x}\) term coefficient.

\(\displaystyle \begin{tabular}{ c| ccc}-13&11&-9&7\\ ~ & ~ &-143& ~ \\ \hline ~ &11& ~ & ~ \end{tabular}\)

Now we add the column up to get.

\(\displaystyle \begin{tabular}{ c| ccc}-13&11&-9&7\\~ & ~ &-143& ~ \\ \hline ~ &11&-152& ~ \end{tabular}\)

Now we multiply the number we got by the zero, and place it under the constant term.

\(\displaystyle \begin{tabular}{ c| ccc}-13&11&-9&7\\ ~ & ~ &-143&1976\\ \hline ~ &11&-152& ~ \end{tabular}\)

Now we add the column together to get.

\(\displaystyle \begin{tabular}{ c| ccc}-13&11&-9&7\\ ~ & ~ &-143&1976\\ \hline ~ &11&-152&1983\end{tabular}\)

The last number is the remainder, so our final answer is \(\displaystyle 1983\).

In order to solve this problem, we need to perform synthetic division. We set up synthetic division by writing down the zero of the expression we are dividing by, and the coefficients of the polynomial on a line.

\(\displaystyle \begin{tabular}{ c| ccc}-7&8&18&7\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}\)

The first step is to bring the coefficient of the \(\displaystyle \uptext{x}^2\) term down.

\(\displaystyle \begin{tabular}{ c| ccc}-7&8&18&7\\ ~ & ~ & ~ & ~ \\ \hline ~ &8& ~ & ~ \end{tabular}\)

Now we multiply the zero by the term we just put down, and place it under the \(\displaystyle \uptext{x}\) term coefficient.

\(\displaystyle \begin{tabular}{ c| ccc}-7&8&18&7\\ ~ & ~ &-56& ~ \\ \hline ~ &8& ~ & ~ \end{tabular}\)

Now we add the column up to get.

\(\displaystyle \begin{tabular}{ c| ccc}-7&8&18&7\\~ & ~ &-56& ~ \\ \hline ~ &8&-38& ~ \end{tabular}\)

Now we multiply the number we got by the zero, and place it under the constant term.

\(\displaystyle \begin{tabular}{ c| ccc}-7&8&18&7\\ ~ & ~ &-56&266\\ \hline ~ &8&-38& ~ \end{tabular}\)

Now we add the column together to get.

\(\displaystyle \begin{tabular}{ c| ccc}-7&8&18&7\\ ~ & ~ &-56&266\\ \hline ~ &8&-38&273\end{tabular}\)

The last number is the remainder, so our final answer is \(\displaystyle 273\).

In order to solve this problem, we need to perform synthetic division. We set up synthetic division by writing down the zero of the expression we are dividing by, and the coefficients of the polynomial on a line.

\(\displaystyle \begin{tabular}{ c| ccc}4&15&-2&-4\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}\)

The first step is to bring the coefficient of the \(\displaystyle \uptext{x}^2\) term down.

\(\displaystyle \begin{tabular}{ c| ccc}4&15&-2&-4\\ ~ & ~ & ~ & ~ \\ \hline ~ &15& ~ & ~ \end{tabular}\)

Now we multiply the zero by the term we just put down, and place it under the \(\displaystyle \uptext{x}\) term coefficient.

\(\displaystyle \begin{tabular}{ c| ccc}4&15&-2&-4\\ ~ & ~ &60& ~ \\ \hline ~ &15& ~ & ~ \end{tabular}\)

Now we add the column up to get.

\(\displaystyle \begin{tabular}{ c| ccc}4&15&-2&-4\\~ & ~ &60& ~ \\ \hline ~ &15&58& ~ \end{tabular}\)

Now we multiply the number we got by the zero, and place it under the constant term.

\(\displaystyle \begin{tabular}{ c| ccc}4&15&-2&-4\\ ~ & ~ &60&232\\ \hline ~ &15&58& ~ \end{tabular}\)

Now we add the column together to get.

\(\displaystyle \begin{tabular}{ c| ccc}4&15&-2&-4\\ ~ & ~ &60&232\\ \hline ~ &15&58&228\end{tabular}\)

The last number is the remainder, so our final answer is \(\displaystyle 228\).

In order to solve this problem, we need to perform synthetic division. We set up synthetic division by writing down the zero of the expression we are dividing by, and the coefficients of the polynomial on a line.

\(\displaystyle \begin{tabular}{ c| ccc}-2&19&-14&5\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}\)

The first step is to bring the coefficient of the \(\displaystyle \uptext{x}^2\) term down.

\(\displaystyle \begin{tabular}{ c| ccc}-2&19&-14&5\\ ~ & ~ & ~ & ~ \\ \hline ~ &19& ~ & ~ \end{tabular}\)

Now we multiply the zero by the term we just put down, and place it under the \(\displaystyle \uptext{x}\) term coefficient.

\(\displaystyle \begin{tabular}{ c| ccc}-2&19&-14&5\\ ~ & ~ &-38& ~ \\ \hline ~ &19& ~ & ~ \end{tabular}\)

Now we add the column up to get.

\(\displaystyle \begin{tabular}{ c| ccc}-2&19&-14&5\\~ & ~ &-38& ~ \\ \hline ~ &19&-52& ~ \end{tabular}\)

Now we multiply the number we got by the zero, and place it under the constant term.

\(\displaystyle \begin{tabular}{ c| ccc}-2&19&-14&5\\ ~ & ~ &-38&104\\ \hline ~ &19&-52& ~ \end{tabular}\)

Now we add the column together to get.

\(\displaystyle \begin{tabular}{ c| ccc}-2&19&-14&5\\ ~ & ~ &-38&104\\ \hline ~ &19&-52&109\end{tabular}\)

The last number is the remainder, so our final answer is \(\displaystyle 109\).

In order to solve this problem, we need to perform synthetic division. We set up synthetic division by writing down the zero of the expression we are dividing by, and the coefficients of the polynomial on a line.

\(\displaystyle \begin{tabular}{ c| ccc}-2&-17&12&-9\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}\)

The first step is to bring the coefficient of the \(\displaystyle \uptext{x}^2\) term down.

\(\displaystyle \begin{tabular}{ c| ccc}-2&-17&12&-9\\ ~ & ~ & ~ & ~ \\ \hline ~ &-17& ~ & ~ \end{tabular}\)

Now we multiply the zero by the term we just put down, and place it under the \(\displaystyle \uptext{x}\) term coefficient.

\(\displaystyle \begin{tabular}{ c| ccc}-2&-17&12&-9\\ ~ & ~ &34& ~ \\ \hline ~ &-17& ~ & ~ \end{tabular}\)

Now we add the column up to get.

\(\displaystyle \begin{tabular}{ c| ccc}-2&-17&12&-9\\~ & ~ &34& ~ \\ \hline ~ &-17&46& ~ \end{tabular}\)

Now we multiply the number we got by the zero, and place it under the constant term.

\(\displaystyle \begin{tabular}{ c| ccc}-2&-17&12&-9\\ ~ & ~ &34&-92\\ \hline ~ &-17&46& ~ \end{tabular}\)

Now we add the column together to get.

\(\displaystyle \begin{tabular}{ c| ccc}-2&-17&12&-9\\ ~ & ~ &34&-92\\ \hline ~ &-17&46&-101\end{tabular}\)

The last number is the remainder, so our final answer is \(\displaystyle -101\).

In order to solve this problem, we need to perform synthetic division. We set up synthetic division by writing down the zero of the expression we are dividing by, and the coefficients of the polynomial on a line.

\(\displaystyle \begin{tabular}{ c| ccc}7&-13&15&2\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}\)

The first step is to bring the coefficient of the \(\displaystyle \uptext{x}^2\) term down.

\(\displaystyle \begin{tabular}{ c| ccc}7&-13&15&2\\ ~ & ~ & ~ & ~ \\ \hline ~ &-13& ~ & ~ \end{tabular}\)

Now we multiply the zero by the term we just put down, and place it under the \(\displaystyle \uptext{x}\) term coefficient.

\(\displaystyle \begin{tabular}{ c| ccc}7&-13&15&2\\ ~ & ~ &-91& ~ \\ \hline ~ &-13& ~ & ~ \end{tabular}\)

Now we add the column up to get.

\(\displaystyle \begin{tabular}{ c| ccc}7&-13&15&2\\~ & ~ &-91& ~ \\ \hline ~ &-13&-76& ~ \end{tabular}\)

Now we multiply the number we got by the zero, and place it under the constant term.

\(\displaystyle \begin{tabular}{ c| ccc}7&-13&15&2\\ ~ & ~ &-91&-532\\ \hline ~ &-13&-76& ~ \end{tabular}\)

Now we add the column together to get.

\(\displaystyle \begin{tabular}{ c| ccc}7&-13&15&2\\ ~ & ~ &-91&-532\\ \hline ~ &-13&-76&-530\end{tabular}\)

The last number is the remainder, so our final answer is \(\displaystyle -530\).

In order to solve this problem, we need to perform synthetic division. We set up synthetic division by writing down the zero of the expression we are dividing by, and the coefficients of the polynomial on a line.

\(\displaystyle \begin{tabular}{ c| ccc}-4&4&-4&5\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}\)

The first step is to bring the coefficient of the \(\displaystyle \uptext{x}^2\) term down.

\(\displaystyle \begin{tabular}{ c| ccc}-4&4&-4&5\\ ~ & ~ & ~ & ~ \\ \hline ~ &4& ~ & ~ \end{tabular}\)

Now we multiply the zero by the term we just put down, and place it under the \(\displaystyle \uptext{x}\) term coefficient.

\(\displaystyle \begin{tabular}{ c| ccc}-4&4&-4&5\\ ~ & ~ &-16& ~ \\ \hline ~ &4& ~ & ~ \end{tabular}\)

Now we add the column up to get.

\(\displaystyle \begin{tabular}{ c| ccc}-4&4&-4&5\\~ & ~ &-16& ~ \\ \hline ~ &4&-20& ~ \end{tabular}\)

Now we multiply the number we got by the zero, and place it under the constant term.

\(\displaystyle \begin{tabular}{ c| ccc}-4&4&-4&5\\ ~ & ~ &-16&80\\ \hline ~ &4&-20& ~ \end{tabular}\)

Now we add the column together to get.

\(\displaystyle \begin{tabular}{ c| ccc}-4&4&-4&5\\ ~ & ~ &-16&80\\ \hline ~ &4&-20&85\end{tabular}\)

The last number is the remainder, so our final answer is \(\displaystyle 85\).

In order to solve this problem, we need to perform synthetic division. We set up synthetic division by writing down the zero of the expression we are dividing by, and the coefficients of the polynomial on a line.

\(\displaystyle \begin{tabular}{ c| ccc}19&6&8&-7\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}\)

The first step is to bring the coefficient of the \(\displaystyle \uptext{x}^2\) term down.

\(\displaystyle \begin{tabular}{ c| ccc}19&6&8&-7\\ ~ & ~ & ~ & ~ \\ \hline ~ &6& ~ & ~ \end{tabular}\)

Now we multiply the zero by the term we just put down, and place it under the \(\displaystyle \uptext{x}\) term coefficient.

\(\displaystyle \begin{tabular}{ c| ccc}19&6&8&-7\\ ~ & ~ &114& ~ \\ \hline ~ &6& ~ & ~ \end{tabular}\)

Now we add the column up to get.

\(\displaystyle \begin{tabular}{ c| ccc}19&6&8&-7\\~ & ~ &114& ~ \\ \hline ~ &6&122& ~ \end{tabular}\)

Now we multiply the number we got by the zero, and place it under the constant term.

\(\displaystyle \begin{tabular}{ c| ccc}19&6&8&-7\\ ~ & ~ &114&2318\\ \hline ~ &6&122& ~ \end{tabular}\)

Now we add the column together to get.

\(\displaystyle \begin{tabular}{ c| ccc}19&6&8&-7\\ ~ & ~ &114&2318\\ \hline ~ &6&122&2311\end{tabular}\)

The last number is the remainder, so our final answer is \(\displaystyle 2311\).

In order to solve this problem, we need to perform synthetic division. We set up synthetic division by writing down the zero of the expression we are dividing by, and the coefficients of the polynomial on a line.

\(\displaystyle \begin{tabular}{ c| ccc}-18&11&4&3\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}\)

The first step is to bring the coefficient of the \(\displaystyle \uptext{x}^2\) term down.

\(\displaystyle \begin{tabular}{ c| ccc}-18&11&4&3\\ ~ & ~ & ~ & ~ \\ \hline ~ &11& ~ & ~ \end{tabular}\)

Now we multiply the zero by the term we just put down, and place it under the \(\displaystyle \uptext{x}\) term coefficient.

\(\displaystyle \begin{tabular}{ c| ccc}-18&11&4&3\\ ~ & ~ &-198& ~ \\ \hline ~ &11& ~ & ~ \end{tabular}\)

Now we add the column up to get.

\(\displaystyle \begin{tabular}{ c| ccc}-18&11&4&3\\~ & ~ &-198& ~ \\ \hline ~ &11&-194& ~ \end{tabular}\)

Now we multiply the number we got by the zero, and place it under the constant term.

\(\displaystyle \begin{tabular}{ c| ccc}-18&11&4&3\\ ~ & ~ &-198&3492\\ \hline ~ &11&-194& ~ \end{tabular}\)

Now we add the column together to get.

\(\displaystyle \begin{tabular}{ c| ccc}-18&11&4&3\\ ~ & ~ &-198&3492\\ \hline ~ &11&-194&3495\end{tabular}\)

The last number is the remainder, so our final answer is \(\displaystyle 3495\).

In order to solve this problem, we need to perform synthetic division. We set up synthetic division by writing down the zero of the expression we are dividing by, and the coefficients of the polynomial on a line.

\(\displaystyle \begin{tabular}{ c| ccc}-15&-8&11&1\\ ~ & ~ & ~ & ~ \\ \hline ~ & ~ & ~ & ~ \end{tabular}\)

The first step is to bring the coefficient of the \(\displaystyle \uptext{x}^2\) term down.

\(\displaystyle \begin{tabular}{ c| ccc}-15&-8&11&1\\ ~ & ~ & ~ & ~ \\ \hline ~ &-8& ~ & ~ \end{tabular}\)

Now we multiply the zero by the term we just put down, and place it under the \(\displaystyle \uptext{x}\) term coefficient.

\(\displaystyle \begin{tabular}{ c| ccc}-15&-8&11&1\\ ~ & ~ &120& ~ \\ \hline ~ &-8& ~ & ~ \end{tabular}\)

Now we add the column up to get.

\(\displaystyle \begin{tabular}{ c| ccc}-15&-8&11&1\\~ & ~ &120& ~ \\ \hline ~ &-8&131& ~ \end{tabular}\)

Now we multiply the number we got by the zero, and place it under the constant term.

\(\displaystyle \begin{tabular}{ c| ccc}-15&-8&11&1\\ ~ & ~ &120&-1965\\ \hline ~ &-8&131& ~ \end{tabular}\)

Now we add the column together to get.

\(\displaystyle \begin{tabular}{ c| ccc}-15&-8&11&1\\ ~ & ~ &120&-1965\\ \hline ~ &-8&131&-1964\end{tabular}\)

The last number is the remainder, so our final answer is \(\displaystyle -1964\).