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Learn Horner's method for efficient polynomial division. Learn this simplified alternative to long division with worked examples and practice.
---
## Understanding Horner's Method
Horner's method is a more efficient and simpler way to perform polynomial division, especially when the divisor is linear, like $$(x-c)$$. This method can be considered a simplification of long division because it only involves writing coefficients and more concise multiplication and addition operations.
Visible text: Horner's method is a more efficient and simpler way to perform polynomial division, especially when the divisor is linear, like . This method can be considered a simplification of long division because it only involves writing coefficients and more concise multiplication and addition operations.
### Comparison with Long Division
To see how Horner's method simplifies the process, let's compare the division of $$P(x) = x^3 - 7x + 8$$ (or $$x^3 + 0x^2 - 7x + 8$$) by $$x-2$$ using both methods:
Visible text: To see how Horner's method simplifies the process, let's compare the division of (or ) by using both methods:
**Long Division:**
```math
\begin{array}{l}
\qquad\quad\space\space x^2 + 2x - 3 \\
x-2\overline{\big)x^3 + 0x^2 - 7x + 8} \\
\quad\space\space \underline{-(x^3 - 2x^2)} \\
\qquad\qquad 2x^2 - 7x \\
\qquad\quad \underline{-(2x^2 - 4x)} \\
\qquad\qquad\qquad -3x + 8 \\
\qquad\qquad\quad \underline{-(-3x + 6)} \\
\qquad\qquad\qquad\qquad\quad 2 \\
\end{array}
```
**Horner's Method:**
```math
\begin{array}{c|cccc}
2 & 1 & 0 & -7 & 8 \\
& & 2 & 4 & -6 \\
\hline
& 1 & 2 & -3 & \boxed{2} \\
\end{array}
```
Notice how Horner's method only focuses on the coefficients $$(1, 0, -7, 8)$$ and the value $$c=2$$. The bottom row in Horner's method $$(1, 2, -3)$$ directly gives the coefficients of the quotient $$1x^2 + 2x - 3$$, and the last number $$\boxed{2}$$ is the remainder.
Visible text: Notice how Horner's method only focuses on the coefficients and the value . The bottom row in Horner's method directly gives the coefficients of the quotient , and the last number is the remainder.
This is much more concise than writing out all the variables and powers as in long division.
This method can only be directly used if the divisor is a linear polynomial of degree one, i.e., in the form $$(x-c)$$ or $$(ax-b)$$ (which can be transformed).
Visible text: This method can only be directly used if the divisor is a linear polynomial of degree one, i.e., in the form or (which can be transformed).
## Preparing to Use Horner's Method
Before performing division with Horner's method, there are several preparation steps:
1. **Identify Coefficients of the Dividend ($$P(x)$$):**
Write down all coefficients of the polynomial to be divided in order, starting from the term with the highest power down to the constant. Ensure no power is skipped. If a term with a certain power is missing, its coefficient is written as $$0$$.
**Example:**
For $$P(x) = 2x^3 + 5x^2 + 6$$, the coefficient of the $$x$$ term is $$0$$. So, the coefficients we write in order are: $$2, 5, 0, 6$$.
2. **Identify the Value $$c$$ from the Divisor ($$x-c$$):**
Determine the value of $$c$$ from the divisor polynomial. Remember, if the divisor is $$x-c$$, the value used is $$c$$. If the divisor is $$x+c$$, it is equivalent to $$x - (-c)$$, so the value used is $$-c$$.
**Example:**
If the divisor is $$x+3$$, then $$x - (-3)$$, so $$c = -3$$.
3. **Set Up the Horner Scheme:**
Draw the Horner scheme or diagram. Place the value $$c$$ on the left and write the coefficients of $$P(x)$$ in the top row.
Visible text: 1. **Identify Coefficients of the Dividend ():**
Write down all coefficients of the polynomial to be divided in order, starting from the term with the highest power down to the constant. Ensure no power is skipped. If a term with a certain power is missing, its coefficient is written as .
**Example:**
For , the coefficient of the term is . So, the coefficients we write in order are: .
2. **Identify the Value from the Divisor ():**
Determine the value of from the divisor polynomial. Remember, if the divisor is , the value used is . If the divisor is , it is equivalent to , so the value used is .
**Example:**
If the divisor is , then , so .
3. **Set Up the Horner Scheme:**
Draw the Horner scheme or diagram. Place the value on the left and write the coefficients of in the top row.
## Division Process with Horner's Method
Here are the steps to perform division using the Horner scheme:
1. **Bring Down the First Coefficient:** Bring down the first coefficient ($$a_n$$) directly to the result row (the bottom row).
2. **Multiply and Place:** Multiply the coefficient just brought down by the value $$c$$. Place the result under the second coefficient ($$a_{n-1}$$).
3. **Add:** Add the second coefficient ($$a_{n-1}$$) to the result from the previous step. Write the sum in the result row, directly below it.
4. **Repeat:** Repeat step $$2$$ (multiply by $$c$$) and step $$3$$ (add to the coefficient above it) for all remaining coefficients.
5. **Final Result:** The last number in the result row is the **remainder ($$S$$)**. The other numbers in the result row, from left to right, are the **coefficients of the quotient polynomial ($$H(x)$$)**, starting from the power $$n-1$$.
Visible text: 1. **Bring Down the First Coefficient:** Bring down the first coefficient () directly to the result row (the bottom row).
2. **Multiply and Place:** Multiply the coefficient just brought down by the value . Place the result under the second coefficient ().
3. **Add:** Add the second coefficient () to the result from the previous step. Write the sum in the result row, directly below it.
4. **Repeat:** Repeat step (multiply by ) and step (add to the coefficient above it) for all remaining coefficients.
5. **Final Result:** The last number in the result row is the **remainder ()**. The other numbers in the result row, from left to right, are the **coefficients of the quotient polynomial ()**, starting from the power .
## Using Horner's Method
Let's divide $$P(x) = 2x^3 + 5x^2 + 6$$ by $$x+3$$ using both methods.
Visible text: Let's divide by using both methods.
**Long Division:**
```math
\begin{array}{l}
\qquad\quad\space 2x^2 - x + 3 \\
x+3\overline{\big)2x^3 + 5x^2 + 0x + 6} \\
\quad\space \underline{-(2x^3 + 6x^2)} \\
\qquad\qquad -x^2 + 0x \\
\qquad\quad \underline{-(-x^2 - 3x)} \\
\qquad\qquad\qquad 3x + 6 \\
\qquad\qquad\quad \underline{-(3x + 9)} \\
\qquad\qquad\qquad\qquad -3 \\
\end{array}
```
**Horner's Method:**
1. **Preparation:**
- Coefficients of $$P(x)$$: $$2, 5, 0$$ (for $$x^1$$), $$6$$ (constant).
- The divisor is $$x+3$$, so $$x - (-3)$$, thus $$c = -3$$.
2. **Horner Process:**
```math
\begin{array}{c|cccc}
-3 & 2 & 5 & 0 & 6 \\
& & -6 & 3 & -9 \\
\hline
& 2 & -1 & 3 & \boxed{-3} \\
\end{array}
```
3. **Scheme Explanation:**
- Bring down the number $$2$$ (coefficient of $$x^3$$).
- $$2 \times (-3) = -6$$. Place $$-6$$ under $$5$$.
- $$5 + (-6) = -1$$. Write $$-1$$ in the result row.
- $$-1 \times (-3) = 3$$. Place $$3$$ under $$0$$.
- $$0 + 3 = 3$$. Write $$3$$ in the result row.
- $$3 \times (-3) = -9$$. Place $$-9$$ under $$6$$.
- $$6 + (-9) = -3$$. Write $$-3$$ (remainder) in the result row on the far right.
4. **Result:**
- The last number in the result row is $$\boxed{-3}$$. This is the **Remainder ($$S$$)**.
- The other numbers are $$2, -1, 3$$. These are the coefficients of the **Quotient ($$H(x)$$)**. Since $$P(x)$$ is degree $$3$$, $$H(x)$$ is degree $$2$$.
- Thus, $$H(x) = 2x^2 - 1x + 3 = 2x^2 - x + 3$$.
5. **Writing in Division Algorithm Form:**
```math
P(x) = (x-c) H(x) + S
```
```math
2x^3 + 5x^2 + 6 = (x+3)(2x^2 - x + 3) - 3
```
Visible text: 1. **Preparation:**
- Coefficients of : (for ), (constant).
- The divisor is , so , thus .
2. **Horner Process:**
3. **Scheme Explanation:**
- Bring down the number (coefficient of ).
- . Place under .
- . Write in the result row.
- . Place under .
- . Write in the result row.
- . Place under .
- . Write (remainder) in the result row on the far right.
4. **Result:**
- The last number in the result row is . This is the **Remainder ()**.
- The other numbers are . These are the coefficients of the **Quotient ()**. Since is degree , is degree .
- Thus, .
5. **Writing in Division Algorithm Form:**
## Exercise
Find the quotient and remainder from the division of $$x^4 + 4$$ by $$x - 1$$ using Horner's method and long division.
Visible text: Find the quotient and remainder from the division of by using Horner's method and long division.
State the result in the form $$P(x) = Q(x) \cdot H(x) + S(x)$$.
Visible text: State the result in the form .
### Answer Key
- Dividend: $$P(x) = x^4 + 4 = x^4 + 0x^3 + 0x^2 + 0x + 4$$.
- Divisor: $$Q(x) = x - 1$$, so $$c = 1$$.
Visible text: - Dividend: .
- Divisor: , so .
**Long Division:**
```math
\begin{array}{l}
\qquad\quad\space\space x^3 + x^2 + x + 1 \\
x-1\overline{\big)x^4 + 0x^3 + 0x^2 + 0x + 4} \\
\quad\space\space \underline{-(x^4 - x^3)} \\
\qquad\qquad x^3 + 0x^2 \\
\qquad\quad \underline{-(x^3 - x^2)} \\
\qquad\qquad\qquad x^2 + 0x \\
\qquad\qquad\quad \underline{-(x^2 - x)} \\
\qquad\qquad\qquad\qquad x + 4 \\
\qquad\qquad\qquad\quad \underline{-(x - 1)} \\
\qquad\qquad\qquad\qquad\qquad 5 \\
\end{array}
```
**Horner's Method:**
```math
\begin{array}{c|ccccc}
1 & 1 & 0 & 0 & 0 & 4 \\
& & 1 & 1 & 1 & 1 \\
\hline
& 1 & 1 & 1 & 1 & \boxed{5} \\
\end{array}
```
**Result:**
- Quotient: $$H(x) = x^3 + x^2 + x + 1$$.
- Remainder: $$S = \boxed{5}$$.
Visible text: - Quotient: .
- Remainder: .
**Writing in Division Algorithm Form:**
```math
x^4 + 4 = (x-1)(x^3 + x^2 + x + 1) + 5
```