📝 Worked Example: The Horner Method
Let's solve a problem step-by-step to see how the numbers move through the "Bagan Horner."
Problem: Find the quotient and remainder when: $$P(x) = x^4 - 3x^2 + 2x - 5$$ is divided by $(x - 2)$.
1. Identify the Coefficients
Before starting, look for "missing" powers of $x$. In this polynomial, there is no $x^3$ term. You must use 0 as a placeholder.
- Coefficients: 1 ($x^4$), 0 ($x^3$), -3 ($x^2$), 2 ($x$), -5 (constant)
- Divisor ($k$): Since we divide by $(x - 2)$, our $k = 2$.
2. The Horner Table Setup
| $x^4$ | $x^3$ | $x^2$ | $x^1$ | $x^0$ | |
|---|---|---|---|---|---|
| 1 | 0 | -3 | 2 | -5 | |
| (2) | ↓ | 2 | 4 | 2 | 8 |
| 1 | 2 | 1 | 4 | (3) |
Step-by-Step Calculation:
- Drop: Bring the first coefficient (1) straight down.
- Multiply & Add: $1 \times 2 = 2$. Add to $0 \rightarrow$ 2.
- Multiply & Add: $2 \times 2 = 4$. Add to $-3 \rightarrow$ 1.
- Multiply & Add: $1 \times 2 = 2$. Add to $2 \rightarrow$ 4.
- Multiply & Add: $4 \times 2 = 8$. Add to $-5 \rightarrow$ 3.
3. Interpreting the Result
The final number in the row is your Remainder, and the preceding numbers are the coefficients of the Quotient.
Important: The degree of the quotient is always one less than the original. Since we started with $x^4$, our result starts with $x^3$.
- Quotient ($Q(x)$): $1x^3 + 2x^2 + 1x + 4$
- Remainder ($R$): $3$
4. Verification (Remainder Theorem)
To ensure the Horner Method worked correctly, we can plug $x=2$ into the original equation: $$P(2) = (2)^4 - 3(2)^2 + 2(2) - 5$$ $$P(2) = 16 - 12 + 4 - 5$$ $$P(2) = 3$$
The results match!
⚠️ Common Pitfall: The "Zero" Placeholder
If you forget to include the 0 for the $x^3$ term, the entire calculation shifts:
- Wrong: $[1, -3, 2, -5]$ ❌
- Correct: $[1, 0, -3, 2, -5]$ ✅
Always double-check that your coefficients represent every power from the highest degree down to the constant.