College

What is the remainder in the synthetic division problem below?

[tex]
-2 \longdiv{1} \begin{array}{llll}
2 & -3 & 1
\end{array}
[/tex]

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

Answer :

To find the remainder using synthetic division, follow these steps:

1. Set up the synthetic division problem:
- The divisor is [tex]\( -2 \)[/tex].
- The polynomial's coefficients are: [tex]\( 2 \)[/tex], [tex]\( -3 \)[/tex], and [tex]\( 1 \)[/tex].

2. Write the coefficients of the polynomial:
- [tex]\( 2 \)[/tex], [tex]\( -3 \)[/tex], and [tex]\( 1 \)[/tex].

3. Perform the synthetic division:
- Bring down the first coefficient, which is [tex]\( 2 \)[/tex].

4. Multiply and add:
- Multiply the brought down number [tex]\( 2 \)[/tex] by the divisor [tex]\( -2 \)[/tex]. This gives [tex]\( -4 \)[/tex].
- Add this result to the next coefficient: [tex]\( -3 + (-4) = -7 \)[/tex].

5. Repeat the process:
- Multiply the result [tex]\( -7 \)[/tex] by [tex]\( -2 \)[/tex]. This gives [tex]\( 14 \)[/tex].
- Add this to the next coefficient: [tex]\( 1 + 14 = 15 \)[/tex].

6. Determine the remainder:
- The last number you get, [tex]\( 15 \)[/tex], is the remainder.

So, the remainder of the division is 15. Unfortunately, none of the options provided (A. 13, B. 7, C. 11, D. 9) match this result. Please double-check the problem setup or the provided options.