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Answer :
To find the remainder of the synthetic division problem where we divide the polynomial [tex]\( P(x) = x^3 + 2x^2 - 3x + 2 \)[/tex] by [tex]\( x + 1 \)[/tex], we can follow these steps:
1. Determine the divisor and the coefficients:
The divisor is [tex]\( x + 1 \)[/tex], which means [tex]\( c = -1 \)[/tex].
The coefficients of the polynomial [tex]\( P(x) \)[/tex] are [tex]\([1, 2, -3, 2]\)[/tex].
2. Setup synthetic division:
Write down the coefficients: [tex]\[ 1, 2, -3, 2 \][/tex]
We will place the value [tex]\( -1 \)[/tex] (the root of the divisor) to the left.
```
-1 | 1 2 -3 2
```
3. Perform the synthetic division steps:
- Bring down the leading coefficient:
```
1
```
- Multiply this value by [tex]\( -1 \)[/tex] and place it below the next coefficient:
```
-1 | 1 2 -3 2
-1
-----------------
1 1
```
- Add the result to the next coefficient:
```
-1 | 1 2 -3 2
-1
-----------------
1 1
```
- Repeat the process: Multiply by [tex]\( -1 \)[/tex] and add:
```
-1 | 1 2 -3 2
-1 -1
-----------------
1 1 -4
```
- Continue with the next term:
```
-1 | 1 2 -3 2
-1 -1 4
-----------------
1 1 -4 6
```
4. Result:
The last number, 6, is the remainder of this division.
So, the remainder of the synthetic division problem is:
[tex]\[ \boxed{6} \][/tex]
1. Determine the divisor and the coefficients:
The divisor is [tex]\( x + 1 \)[/tex], which means [tex]\( c = -1 \)[/tex].
The coefficients of the polynomial [tex]\( P(x) \)[/tex] are [tex]\([1, 2, -3, 2]\)[/tex].
2. Setup synthetic division:
Write down the coefficients: [tex]\[ 1, 2, -3, 2 \][/tex]
We will place the value [tex]\( -1 \)[/tex] (the root of the divisor) to the left.
```
-1 | 1 2 -3 2
```
3. Perform the synthetic division steps:
- Bring down the leading coefficient:
```
1
```
- Multiply this value by [tex]\( -1 \)[/tex] and place it below the next coefficient:
```
-1 | 1 2 -3 2
-1
-----------------
1 1
```
- Add the result to the next coefficient:
```
-1 | 1 2 -3 2
-1
-----------------
1 1
```
- Repeat the process: Multiply by [tex]\( -1 \)[/tex] and add:
```
-1 | 1 2 -3 2
-1 -1
-----------------
1 1 -4
```
- Continue with the next term:
```
-1 | 1 2 -3 2
-1 -1 4
-----------------
1 1 -4 6
```
4. Result:
The last number, 6, is the remainder of this division.
So, the remainder of the synthetic division problem is:
[tex]\[ \boxed{6} \][/tex]
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