Answer :
Sure, let's divide the polynomial [tex]\( 4x^5 - 5x^3 + 6x^2 - 7x + 2 \)[/tex] by [tex]\( x - 2 \)[/tex] using synthetic division. Here are the steps involved:
1. Setup:
- Write down the coefficients of the dividend polynomial: [tex]\(4, 0, -5, 6, -7, 2\)[/tex]. Note that the coefficient for [tex]\(x^4\)[/tex] is 0 because that term is missing in the original polynomial.
- The divisor is [tex]\( x - 2 \)[/tex], so the root used in synthetic division is [tex]\( 2 \)[/tex].
2. Perform Synthetic Division:
- Write the root [tex]\( 2 \)[/tex] on the left.
- Write the coefficients [tex]\(4, 0, -5, 6, -7, 2\)[/tex] in a row.
- Bring down the first coefficient [tex]\(4\)[/tex] directly below the row of coefficients. This will be the first coefficient of the quotient.
3. Iterate Through the Coefficients:
- Multiply the root [tex]\(2\)[/tex] by the current value (initially the first coefficient [tex]\(4\)[/tex]), and add the product to the next coefficient.
- Repeat the process for each coefficient.
Let's show these steps explicitly:
### Steps in Synthetic Division:
| | 2 | 4 | 0 | -5 | 6 | -7 | 2 |
|---|---|---|---|----|---|----|--|
| | | 4 | 8 | 11 | 28| 49 | 100 |
| |---|---|---|----|---|----|----|
| = | 4 | 8 | 11| 28 | 49 | 100|
1. Bring down the [tex]\( 4 \)[/tex]:
[tex]\[ \begin{array}{c|ccccc} 2 & 4 & 0 & -5 & 6 & -7 & 2 \\ \hline & 4 & & & & & \\ \end{array} \][/tex]
2. Multiply [tex]\(4\)[/tex] by [tex]\(2\)[/tex] and add to the next coefficient [tex]\(0\)[/tex]:
[tex]\[ \begin{array}{c|ccccc} 2 & 4 & 0 & -5 & 6 & -7 & 2 \\ \hline & 4 & 8 & & & & \\ \end{array} \][/tex]
3. Multiply [tex]\(8\)[/tex] by [tex]\(2\)[/tex] and add to the next coefficient [tex]\(-5\)[/tex]:
[tex]\[ \begin{array}{c|ccccc} 2 & 4 & 0 & -5 & 6 & -7 & 2 \\ \hline & 4 & 8 & 11 & & & \\ \end{array} \][/tex]
4. Multiply [tex]\(11\)[/tex] by [tex]\(2\)[/tex] and add to the next coefficient [tex]\(6\)[/tex]:
[tex]\[ \begin{array}{c|ccccc} 2 & 4 & 0 & -5 & 6 & -7 & 2 \\ \hline & 4 & 8 & 11 & 28 & & \\ \end{array} \][/tex]
5. Multiply [tex]\(28\)[/tex] by [tex]\(2\)[/tex] and add to the next coefficient [tex]\(-7\)[/tex]:
[tex]\[ \begin{array}{c|ccccc} 2 & 4 & 0 & -5 & 6 & -7 & 2 \\ \hline & 4 & 8 & 11 & 28 & 49 & \\ \end{array} \][/tex]
6. Finally, multiply [tex]\(49\)[/tex] by [tex]\(2\)[/tex] and add to the next coefficient [tex]\(2\)[/tex]:
[tex]\[ \begin{array}{c|ccccc} 2 & 4 & 0 & -5 & 6 & -7 & 2 \\ \hline & 4 & 8 & 11 & 28 & 49 & 100 \\ \end{array} \][/tex]
### Conclusion:
- The quotient polynomial is: [tex]\(4x^4 + 8x^3 + 11x^2 + 28x + 49\)[/tex]
- The remainder is: [tex]\(100\)[/tex]
Thus, when [tex]\( 4x^5 - 5x^3 + 6x^2 - 7x + 2 \)[/tex] is divided by [tex]\( x - 2 \)[/tex], we get:
[tex]\[ \frac{4x^5 - 5x^3 + 6x^2 - 7x + 2}{x-2} = 4x^4 + 8x^3 + 11x^2 + 28x + 49 \text{ with a remainder of } 100. \][/tex]
1. Setup:
- Write down the coefficients of the dividend polynomial: [tex]\(4, 0, -5, 6, -7, 2\)[/tex]. Note that the coefficient for [tex]\(x^4\)[/tex] is 0 because that term is missing in the original polynomial.
- The divisor is [tex]\( x - 2 \)[/tex], so the root used in synthetic division is [tex]\( 2 \)[/tex].
2. Perform Synthetic Division:
- Write the root [tex]\( 2 \)[/tex] on the left.
- Write the coefficients [tex]\(4, 0, -5, 6, -7, 2\)[/tex] in a row.
- Bring down the first coefficient [tex]\(4\)[/tex] directly below the row of coefficients. This will be the first coefficient of the quotient.
3. Iterate Through the Coefficients:
- Multiply the root [tex]\(2\)[/tex] by the current value (initially the first coefficient [tex]\(4\)[/tex]), and add the product to the next coefficient.
- Repeat the process for each coefficient.
Let's show these steps explicitly:
### Steps in Synthetic Division:
| | 2 | 4 | 0 | -5 | 6 | -7 | 2 |
|---|---|---|---|----|---|----|--|
| | | 4 | 8 | 11 | 28| 49 | 100 |
| |---|---|---|----|---|----|----|
| = | 4 | 8 | 11| 28 | 49 | 100|
1. Bring down the [tex]\( 4 \)[/tex]:
[tex]\[ \begin{array}{c|ccccc} 2 & 4 & 0 & -5 & 6 & -7 & 2 \\ \hline & 4 & & & & & \\ \end{array} \][/tex]
2. Multiply [tex]\(4\)[/tex] by [tex]\(2\)[/tex] and add to the next coefficient [tex]\(0\)[/tex]:
[tex]\[ \begin{array}{c|ccccc} 2 & 4 & 0 & -5 & 6 & -7 & 2 \\ \hline & 4 & 8 & & & & \\ \end{array} \][/tex]
3. Multiply [tex]\(8\)[/tex] by [tex]\(2\)[/tex] and add to the next coefficient [tex]\(-5\)[/tex]:
[tex]\[ \begin{array}{c|ccccc} 2 & 4 & 0 & -5 & 6 & -7 & 2 \\ \hline & 4 & 8 & 11 & & & \\ \end{array} \][/tex]
4. Multiply [tex]\(11\)[/tex] by [tex]\(2\)[/tex] and add to the next coefficient [tex]\(6\)[/tex]:
[tex]\[ \begin{array}{c|ccccc} 2 & 4 & 0 & -5 & 6 & -7 & 2 \\ \hline & 4 & 8 & 11 & 28 & & \\ \end{array} \][/tex]
5. Multiply [tex]\(28\)[/tex] by [tex]\(2\)[/tex] and add to the next coefficient [tex]\(-7\)[/tex]:
[tex]\[ \begin{array}{c|ccccc} 2 & 4 & 0 & -5 & 6 & -7 & 2 \\ \hline & 4 & 8 & 11 & 28 & 49 & \\ \end{array} \][/tex]
6. Finally, multiply [tex]\(49\)[/tex] by [tex]\(2\)[/tex] and add to the next coefficient [tex]\(2\)[/tex]:
[tex]\[ \begin{array}{c|ccccc} 2 & 4 & 0 & -5 & 6 & -7 & 2 \\ \hline & 4 & 8 & 11 & 28 & 49 & 100 \\ \end{array} \][/tex]
### Conclusion:
- The quotient polynomial is: [tex]\(4x^4 + 8x^3 + 11x^2 + 28x + 49\)[/tex]
- The remainder is: [tex]\(100\)[/tex]
Thus, when [tex]\( 4x^5 - 5x^3 + 6x^2 - 7x + 2 \)[/tex] is divided by [tex]\( x - 2 \)[/tex], we get:
[tex]\[ \frac{4x^5 - 5x^3 + 6x^2 - 7x + 2}{x-2} = 4x^4 + 8x^3 + 11x^2 + 28x + 49 \text{ with a remainder of } 100. \][/tex]
