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To solve the problem using synthetic division, we divide the polynomial \( x^5 - x^4 + 9x^3 - 9x^2 + x - 8 \) by \( x - 1 \).

### Step-by-step process:

1. **Set up synthetic division**: The divisor is \( x - 1 \), which means the root \( x = 1 \). This will be the number used in synthetic division.

2. **List coefficients**: Write down the coefficients of the polynomial in order:
   \[
   [1, -1, 9, -9, 1, -8]
   \]
   These correspond to the terms \( x^5, x^4, x^3, x^2, x, \text{constant} \).

3. **Perform synthetic division**:
   - Bring down the first coefficient \( 1 \).
   - Multiply \( 1 \) (root) by \( 1 \) (brought down coefficient), and add the result to the next coefficient \( -1 \).
   - Continue this process across all terms.

   Let's perform the calculations now.The result of the synthetic division is:

- **Quotient**: \( x^4 + 9x^2 + 1 \)
- **Remainder**: \( -7 \)

So the division of \( x^5 - x^4 + 9x^3 - 9x^2 + x - 8 \) by \( x - 1 \) results in:
\[
\frac{x^5 - x^4 + 9x^3 - 9x^2 + x - 8}{x - 1} = x^4 + 9x^2 + 1 - \frac{7}{x - 1}
\]

Would you like a detailed breakdown of the steps or have any further questions?

Here are five related questions for further exploration:
1. What changes in synthetic division if the divisor is \( x + 1 \) instead of \( x - 1 \)?
2. How would you check the result of synthetic division using long division?
3. What is the relationship between synthetic division and the Remainder Theorem?
4. Can synthetic division be used when dividing by a quadratic expression?
5. What happens if the remainder is zero in a synthetic division problem?

**Tip**: Synthetic division only works for divisors of the form \( x - c \), where \( c \) is a constant.

Find the quotient and remainder using synthetic division for (x^5 - x^4 + 9x^3 - 9x^2 + x - 8) / (x - 1).

Learn how to use synthetic division to find the quotient and remainder of (x^5 - x^4 + 9x^3 - 9x^2 + x - 8) divided by (x - 1). This problem demonstrates key algebraic concepts and the Remainder Theorem, suitable for students in Grades 10-12.

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