Question

use synthetic division to find the remainder.\\(dfrac{6x^{4}+ 11x^{3}- 16x^{2}- 11x + 10}{x - 1}\\)

Explanation:

Step1: Identify the root

For the divisor \(x - 1\), the root \(r\) is \(1\) (since \(x - 1=0\) gives \(x = 1\)).

Step2: Set up synthetic division

Write the coefficients of the polynomial \(6x^{4}+11x^{3}-16x^{2}-11x + 10\): \(6\), \(11\), \(-16\), \(-11\), \(10\).
Set up the synthetic division as follows:
Bring down the first coefficient (\(6\)):
\[

$$\begin{array}{r|rrrrr} 1 & 6 & 11 & -16 & -11 & 10 \\ & & & & & \\ \hline & 6 & & & & \\ \end{array}$$

\]

Step3: Multiply and add

Multiply the brought - down value (\(6\)) by the root (\(1\)): \(6\times1 = 6\). Add this to the next coefficient (\(11\)): \(11 + 6=17\).
\[

$$\begin{array}{r|rrrrr} 1 & 6 & 11 & -16 & -11 & 10 \\ & & 6 & & & \\ \hline & 6 & 17 & & & \\ \end{array}$$

\]
Multiply \(17\) by \(1\): \(17\times1 = 17\). Add this to the next coefficient (\(-16\)): \(-16+17 = 1\).
\[

$$\begin{array}{r|rrrrr} 1 & 6 & 11 & -16 & -11 & 10 \\ & & 6 & 17 & & \\ \hline & 6 & 17 & 1 & & \\ \end{array}$$

\]
Multiply \(1\) by \(1\): \(1\times1 = 1\). Add this to the next coefficient (\(-11\)): \(-11 + 1=-10\).
\[

$$\begin{array}{r|rrrrr} 1 & 6 & 11 & -16 & -11 & 10 \\ & & 6 & 17 & 1 & \\ \hline & 6 & 17 & 1 & -10 & \\ \end{array}$$

\]
Multiply \(-10\) by \(1\): \(-10\times1=-10\). Add this to the last coefficient (\(10\)): \(10+( - 10)=0\).
\[

$$\begin{array}{r|rrrrr} 1 & 6 & 11 & -16 & -11 & 10 \\ & & 6 & 17 & 1 & -10 \\ \hline & 6 & 17 & 1 & -10 & 0 \\ \end{array}$$

\]
The last number in the bottom row is the remainder.

Answer:

\(0\)