Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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A factor is a number that divides evenly into another number. If long division results in a remainder of 0, then the division must have come out evenly and both the quotient and the divisor must be factors of the dividend.
Find the quotient for 14x^2+31x+6/2x+5 using long division. Indicate the quotient with any remainders. Example attached.
![The image shows a long division of polynomials and additional explanations.
1. **Polynomial Division:**
The division problem is \((3x^3 + 4x^2 + 7x + 22) \div (x + 2)\).
- **Step 1:** Divide the first term of the dividend by the first term of the divisor:
\[\frac{3x^3}{x} = 3x^2.\]
Write \(3x^2\) above the division line.
- **Step 2:** Multiply \(3x^2\) by the entire divisor \(x + 2\), resulting in \(3x^3 + 6x^2\).
Subtract this from the original polynomial to get \(-2x^2 + 7x\).
- **Step 3:** Repeat the process with the new polynomial.
Divide \(-2x^2\) by \(x\):
\[\frac{-2x^2}{x} = -2x.\]
Multiply and subtract to get \(11x + 22\).
- **Step 4:** Continue the process:
\[\frac{11x}{x} = 11.\]
Multiply and subtract to achieve a remainder of 0.
The quotient is \(3x^2 - 2x + 11\).
2. **Conclusion:**
- The final statement concludes that \(x + 2\) is a factor of the original polynomial.
3. **Explanations:**
On the right side:
1. \(\frac{3x^3}{x} = 3x^2\) is the initial step to find the first term of the quotient.
2. \(\frac{-2x^2}{x} = -2x\) simplifies to find the second term of the quotient.
3. \(\frac{11x}{x} = 11\) determines the final term.
Overall, the image details how polynomial long division results in a quotient and confirms that the divisor is a factor when the remainder is zero.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F74c338b6-3b73-4340-9312-2ab60a3dd824%2Fba6e12b7-8ebb-40ac-9c99-c510eb802a76%2Fmdj7lsf_processed.png&w=3840&q=75)
Transcribed Image Text:The image shows a long division of polynomials and additional explanations.
1. **Polynomial Division:**
The division problem is \((3x^3 + 4x^2 + 7x + 22) \div (x + 2)\).
- **Step 1:** Divide the first term of the dividend by the first term of the divisor:
\[\frac{3x^3}{x} = 3x^2.\]
Write \(3x^2\) above the division line.
- **Step 2:** Multiply \(3x^2\) by the entire divisor \(x + 2\), resulting in \(3x^3 + 6x^2\).
Subtract this from the original polynomial to get \(-2x^2 + 7x\).
- **Step 3:** Repeat the process with the new polynomial.
Divide \(-2x^2\) by \(x\):
\[\frac{-2x^2}{x} = -2x.\]
Multiply and subtract to get \(11x + 22\).
- **Step 4:** Continue the process:
\[\frac{11x}{x} = 11.\]
Multiply and subtract to achieve a remainder of 0.
The quotient is \(3x^2 - 2x + 11\).
2. **Conclusion:**
- The final statement concludes that \(x + 2\) is a factor of the original polynomial.
3. **Explanations:**
On the right side:
1. \(\frac{3x^3}{x} = 3x^2\) is the initial step to find the first term of the quotient.
2. \(\frac{-2x^2}{x} = -2x\) simplifies to find the second term of the quotient.
3. \(\frac{11x}{x} = 11\) determines the final term.
Overall, the image details how polynomial long division results in a quotient and confirms that the divisor is a factor when the remainder is zero.
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