Find the remainder when `x^(6)-4x^(5)+3x^(4)-2x^(2)+x-1` is divided with `x+2`

Answer

To find the remainder when the polynomial \( P(x) = x^6 - 4x^5 + 3x^4 - 2x^2 + x - 1 \) is divided by \( x + 2 \), we can use the Remainder Theorem. According to the theorem, the remainder of the division of a polynomial \( P(x) \) by \( x - a \) is \( P(a) \). In our case, since we are dividing by \( x + 2 \), we set \( a = -2 \).

### Step-by-Step Solution:

1. **Identify the Polynomial**:
\[ P(x) = x^6 - 4x^5 + 3x^4 - 2x^2 + x - 1 \]

2. **Substitute \( x = -2 \)**:
We need to calculate \( P(-2) \):
\[ P(-2) = (-2)^6 - 4(-2)^5 + 3(-2)^4 - 2(-2)^2 + (-2) - 1 \]

3. **Calculate Each Term**:
- Calculate \( (-2)^6 \):
\[ (-2)^6 = 64 \]
- Calculate \( -4(-2)^5 \):
\[ -4(-2)^5 = -4 \times -32 = 128 \]
- Calculate \( 3(-2)^4 \):
\[ 3(-2)^4 = 3 \times 16 = 48 \]
- Calculate \( -2(-2)^2 \):
\[ -2(-2)^2 = -2 \times 4 = -8 \]
- Calculate \( -2 \):
\[ -2 = -2 \]
- Calculate \( -1 \):
\[ -1 = -1 \]

4. **Combine All the Terms**:
Now, we sum all the calculated values:
\[ P(-2) = 64 + 128 + 48 - 8 - 2 - 1 \]
- First, combine \( 64 + 128 = 192 \)
- Then, \( 192 + 48 = 240 \)
- Next, \( 240 - 8 = 232 \)
- Then, \( 232 - 2 = 230 \)
- Finally, \( 230 - 1 = 229 \)

5. **Conclusion**:
The remainder when \( P(x) \) is divided by \( x + 2 \) is:
\[ \text{Remainder} = 229 \]