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

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Accepted Answer

# 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:
$0$

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QUESTION IMAGE

Question

Explanation:

Step1: Identify the root

Step2: Set up synthetic division

Step3: Multiply and add

Answer: