Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter8: Polynomials
Section8.1: Adding And Subtracting Polynomials
Problem 59PPS
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Question
![**Dividing Polynomials: Understanding Quotient and Remainder**
In this example, we will learn how to divide polynomials and express the answer in the form:
\[ \text{quotient} + \frac{\text{remainder}}{\text{divisor}} \]
Let's solve the following division problem:
\[ \frac{x^3 - 7x^2 + 40}{x - 3} \]
Here, we need to divide \( x^3 - 7x^2 + 40 \) by \( x - 3 \).
To proceed:
1. **Divide the first term of the numerator by the first term of the denominator:**
The first term of the numerator is \( x^3 \) and the first term of the denominator is \( x \).
So, \[ \frac{x^3}{x} = x^2 \]
2. **Multiply the entire divisor by this result ( \( x^2 \) ) and subtract from the original polynomial:**
\[ (x - 3) \cdot x^2 = x^3 - 3x^2 \]
Subtract this product from the original polynomial:
\[ (x^3 - 7x^2 + 40) - (x^3 - 3x^2) = -4x^2 + 40 \]
3. **Repeat the process with the new polynomial \( -4x^2 + 40 \):**
Divide the first term \( -4x^2 \) by the first term of the divisor \( x \):
\[ \frac{-4x^2}{x} = -4x \]
Multiply the entire divisor by this result ( \( -4x \) ) and subtract:
\[ (x - 3) \cdot (-4x) = -4x^2 + 12x \]
Subtract this product from \( -4x^2 + 40 \):
\[ (-4x^2 + 40) - (-4x^2 + 12x) = -12x + 40 \]
4. **Continue with the term \( -12x + 40 \):**
Divide the first term \( -12x \) by the first term of the divisor \( x \):
\[ \frac{-12x}{x} =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd2cfdfc8-ec6f-4853-bb37-f02ea76b87be%2Fcc21b4b4-7e5d-449a-abcd-6a6701b33f46%2Fe7potd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Dividing Polynomials: Understanding Quotient and Remainder**
In this example, we will learn how to divide polynomials and express the answer in the form:
\[ \text{quotient} + \frac{\text{remainder}}{\text{divisor}} \]
Let's solve the following division problem:
\[ \frac{x^3 - 7x^2 + 40}{x - 3} \]
Here, we need to divide \( x^3 - 7x^2 + 40 \) by \( x - 3 \).
To proceed:
1. **Divide the first term of the numerator by the first term of the denominator:**
The first term of the numerator is \( x^3 \) and the first term of the denominator is \( x \).
So, \[ \frac{x^3}{x} = x^2 \]
2. **Multiply the entire divisor by this result ( \( x^2 \) ) and subtract from the original polynomial:**
\[ (x - 3) \cdot x^2 = x^3 - 3x^2 \]
Subtract this product from the original polynomial:
\[ (x^3 - 7x^2 + 40) - (x^3 - 3x^2) = -4x^2 + 40 \]
3. **Repeat the process with the new polynomial \( -4x^2 + 40 \):**
Divide the first term \( -4x^2 \) by the first term of the divisor \( x \):
\[ \frac{-4x^2}{x} = -4x \]
Multiply the entire divisor by this result ( \( -4x \) ) and subtract:
\[ (x - 3) \cdot (-4x) = -4x^2 + 12x \]
Subtract this product from \( -4x^2 + 40 \):
\[ (-4x^2 + 40) - (-4x^2 + 12x) = -12x + 40 \]
4. **Continue with the term \( -12x + 40 \):**
Divide the first term \( -12x \) by the first term of the divisor \( x \):
\[ \frac{-12x}{x} =
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