Cameron wants to calculate 23 x 23. She says, 20 × 20 = 400 and 3x3=9 SO 23 x 23 400+ 9 = 409 Critique Cameron's reasoning and compare her work with the steps in the partial-products method for calculating 23 x 23.
Cameron wants to calculate 23 x 23. She says, 20 × 20 = 400 and 3x3=9 SO 23 x 23 400+ 9 = 409 Critique Cameron's reasoning and compare her work with the steps in the partial-products method for calculating 23 x 23.
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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![### Analyzing Multiplication Using the Partial-Products Method
**Problem Statement:**
Cameron wants to calculate \( 23 \times 23 \). She says:
\[
20 \times 20 = 400 \text{ and } 3 \times 3 = 9
\]
\[
\text{so}
\]
\[
23 \times 23 = 400 + 9 = 409
\]
**Task:**
Critique Cameron's reasoning and compare her work with the steps in the partial-products method for calculating \( 23 \times 23 \).
### Solution Analysis
Cameron's approach to calculating \( 23 \times 23 \) involves breaking down the numbers into their tens and units components. She performs the following steps:
- Calculates \( 20 \times 20 = 400 \)
- Calculates \( 3 \times 3 = 9 \)
- Adds the results together: \( 400 + 9 = 409 \)
However, there is an error in Cameron's method. This method incorrectly misses out on the cross-products that arise when multiplying the tens and units parts together.
### Correct Partial-Products Method:
The correct method involves the following steps:
1. **Break Down Each Number:**
Break down 23 into 20 and 3.
2. **Calculate Partial Products:**
- Multiply each part of the first number by each part of the second number.
\[
23 \times 23 = (20 + 3) \times (20 + 3)
\]
Expand this product using the distributive property:
\[
= (20 \times 20) + (20 \times 3) + (3 \times 20) + (3 \times 3)
\]
3. **Perform Each Multiplication:**
- \( 20 \times 20 = 400 \)
- \( 20 \times 3 = 60 \)
- \( 3 \times 20 = 60 \)
- \( 3 \times 3 = 9 \)
4. **Add All Partial Products Together:**
\[
400 + 60 + 60 + 9 = 529
\]
### Conclusion:
- Cameron's approach is incorrect because she only considered the squares of the tens and units without considering the cross products.
- The](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8e4d9a92-7f6d-41e2-a11d-6abc746a28c0%2Fed83385c-810c-4087-a6a5-04a20571d9eb%2F8xvgtrx_processed.png&w=3840&q=75)
Transcribed Image Text:### Analyzing Multiplication Using the Partial-Products Method
**Problem Statement:**
Cameron wants to calculate \( 23 \times 23 \). She says:
\[
20 \times 20 = 400 \text{ and } 3 \times 3 = 9
\]
\[
\text{so}
\]
\[
23 \times 23 = 400 + 9 = 409
\]
**Task:**
Critique Cameron's reasoning and compare her work with the steps in the partial-products method for calculating \( 23 \times 23 \).
### Solution Analysis
Cameron's approach to calculating \( 23 \times 23 \) involves breaking down the numbers into their tens and units components. She performs the following steps:
- Calculates \( 20 \times 20 = 400 \)
- Calculates \( 3 \times 3 = 9 \)
- Adds the results together: \( 400 + 9 = 409 \)
However, there is an error in Cameron's method. This method incorrectly misses out on the cross-products that arise when multiplying the tens and units parts together.
### Correct Partial-Products Method:
The correct method involves the following steps:
1. **Break Down Each Number:**
Break down 23 into 20 and 3.
2. **Calculate Partial Products:**
- Multiply each part of the first number by each part of the second number.
\[
23 \times 23 = (20 + 3) \times (20 + 3)
\]
Expand this product using the distributive property:
\[
= (20 \times 20) + (20 \times 3) + (3 \times 20) + (3 \times 3)
\]
3. **Perform Each Multiplication:**
- \( 20 \times 20 = 400 \)
- \( 20 \times 3 = 60 \)
- \( 3 \times 20 = 60 \)
- \( 3 \times 3 = 9 \)
4. **Add All Partial Products Together:**
\[
400 + 60 + 60 + 9 = 529
\]
### Conclusion:
- Cameron's approach is incorrect because she only considered the squares of the tens and units without considering the cross products.
- The
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