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------------------------------------------------ What is the remainder in the synthetic division problem below?

[tex]
\[
\begin{array}{c|cccc}
-2 & 1 & 2 & -3 & 1 \\
\end{array}
\]
[/tex]

A. 9
B. 7
C. 11
D. 13

We want to find the remainder when the polynomial
[tex]$$
P(x) = x^3 + 2x^2 - 3x + 1
$$[/tex]
is divided by [tex]$(x + 2)$[/tex]. Since the divisor is written as [tex]$x - r$[/tex], we have [tex]$r = -2$[/tex].

We perform synthetic division using the coefficients of [tex]$P(x)$[/tex]:
[tex]$$
1,\quad 2,\quad -3,\quad 1.
$$[/tex]

The steps are as follows:

1. Write the number [tex]$-2$[/tex] (our [tex]$r$[/tex] value) to the left and list the coefficients:
[tex]$$
\begin{array}{cccc}
-2 & \vline & 1 & 2 & -3 & 1 \\
\end{array}
$$[/tex]

2. Bring down the first coefficient ([tex]$1$[/tex]) directly:
[tex]$$
b_0 = 1.
$$[/tex]

3. Multiply [tex]$b_0$[/tex] by [tex]$-2$[/tex]:
[tex]$$
1 \times (-2) = -2.
$$[/tex]
Add this result to the next coefficient ([tex]$2$[/tex]):
[tex]$$
2 + (-2) = 0.
$$[/tex]
So,
[tex]$$
b_1 = 0.
$$[/tex]

4. Multiply [tex]$b_1$[/tex] by [tex]$-2$[/tex]:
[tex]$$
0 \times (-2) = 0.
$$[/tex]
Add this to the next coefficient ([tex]$-3$[/tex]):
[tex]$$
-3 + 0 = -3.
$$[/tex]
Thus,
[tex]$$
b_2 = -3.
$$[/tex]

5. Multiply [tex]$b_2$[/tex] by [tex]$-2$[/tex]:
[tex]$$
(-3) \times (-2) = 6.
$$[/tex]
Add this last product to the final coefficient ([tex]$1$[/tex]):
[tex]$$
1 + 6 = 7.
$$[/tex]
The resulting number is the remainder:
[tex]$$
\text{Remainder} = 7.
$$[/tex]

Thus, the remainder of the synthetic division is [tex]$\boxed{7}$[/tex].