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What is the remainder in the synthetic division problem below?

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

A. 3
B. 4
C. 5
D. 6

Answer :

To find the remainder in the synthetic division problem, we need to follow the steps of synthetic division using the given coefficients and constant.

1. Set Up Synthetic Division:
- We're dividing the polynomial represented by its coefficients: [tex]\(1\)[/tex], [tex]\(2\)[/tex], [tex]\(-3\)[/tex].
- The divisor in synthetic division is [tex]\(1\)[/tex].

2. Synthetic Division Steps:
- Bring down the first coefficient: [tex]\(1\)[/tex].
- Multiply this coefficient by the divisor ([tex]\(3\)[/tex]): [tex]\(1 \times 3 = 3\)[/tex].
- Add this result to the next coefficient: [tex]\(2 + 3 = 5\)[/tex].
- Repeat the process: Multiply [tex]\(5\)[/tex] by the divisor: [tex]\(5 \times 3 = 15\)[/tex].
- Add this result to the next coefficient: [tex]\(-3 + 15 = 12\)[/tex].

3. Conclusion:
- The final result, [tex]\(12\)[/tex], is the remainder of the division.

The remainder in this synthetic division problem is [tex]\(12\)[/tex]. Therefore, none of the given options (A. 3, B. 4, C. 5, D. 6) are correct according to this calculation.

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