5) Use long division to find the quotient Q(x) and remainder R(x). P(x) = 6x³ 2x² + 4x - 1 and d(x) = x - - 3 -
5) Use long division to find the quotient Q(x) and remainder R(x). P(x) = 6x³ 2x² + 4x - 1 and d(x) = x - - 3 -
Advanced Engineering Mathematics
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Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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![**Problem 5: Long Division of Polynomials**
Use long division to find the quotient \( Q(x) \) and remainder \( R(x) \).
Given polynomials:
\[ P(x) = 6x^3 - 2x^2 + 4x - 1 \]
\[ d(x) = x - 3 \]
### Steps:
1. **Divide the first term of the dividend by the first term of the divisor:**
- \(\frac{6x^3}{x} = 6x^2\)
2. **Multiply the entire divisor by this result:**
- \(6x^2 \times (x - 3) = 6x^3 - 18x^2\)
3. **Subtract the result from the original polynomial:**
- \((6x^3 - 2x^2 + 4x - 1) - (6x^3 - 18x^2) = 16x^2 + 4x - 1\)
4. **Repeat the process with the new polynomial:**
- \(\frac{16x^2}{x} = 16x\)
- \(16x \times (x - 3) = 16x^2 - 48x\)
- \((16x^2 + 4x - 1) - (16x^2 - 48x) = 52x - 1\)
5. **Continue division:**
- \(\frac{52x}{x} = 52\)
- \(52 \times (x - 3) = 52x - 156\)
- \((52x - 1) - (52x - 156) = 155\)
The final quotient is \( Q(x) = 6x^2 + 16x + 52 \) and the remainder is \( R(x) = 155 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1b897147-dac4-49d3-a0b7-87acbe9c344d%2F84c46c8b-9bf7-47f0-8030-9ea352f80b40%2Fpd3n6te_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 5: Long Division of Polynomials**
Use long division to find the quotient \( Q(x) \) and remainder \( R(x) \).
Given polynomials:
\[ P(x) = 6x^3 - 2x^2 + 4x - 1 \]
\[ d(x) = x - 3 \]
### Steps:
1. **Divide the first term of the dividend by the first term of the divisor:**
- \(\frac{6x^3}{x} = 6x^2\)
2. **Multiply the entire divisor by this result:**
- \(6x^2 \times (x - 3) = 6x^3 - 18x^2\)
3. **Subtract the result from the original polynomial:**
- \((6x^3 - 2x^2 + 4x - 1) - (6x^3 - 18x^2) = 16x^2 + 4x - 1\)
4. **Repeat the process with the new polynomial:**
- \(\frac{16x^2}{x} = 16x\)
- \(16x \times (x - 3) = 16x^2 - 48x\)
- \((16x^2 + 4x - 1) - (16x^2 - 48x) = 52x - 1\)
5. **Continue division:**
- \(\frac{52x}{x} = 52\)
- \(52 \times (x - 3) = 52x - 156\)
- \((52x - 1) - (52x - 156) = 155\)
The final quotient is \( Q(x) = 6x^2 + 16x + 52 \) and the remainder is \( R(x) = 155 \).
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