Answer :
Let's analyze Clara's work step-by-step and find if any mistakes were made in multiplying the binomials [tex]\((3 - 6y^2)\)[/tex] and [tex]\((y^2 + 2)\)[/tex].
### Correct Method Using the Distributive Property
To multiply the binomials, we should use the distributive property to distribute each term inside the first parenthesis to each term inside the second parenthesis. The distributive property states:
[tex]\[ (a + b)(c + d) = ac + ad + bc + bd \][/tex]
For our given binomials, [tex]\((3 - 6y^2)\)[/tex] and [tex]\((y^2 + 2)\)[/tex], we perform the following steps:
1. Distribute [tex]\(3\)[/tex]:
[tex]\[ 3 \cdot y^2 + 3 \cdot 2 \][/tex]
This gives us:
[tex]\[ 3y^2 + 6 \][/tex]
2. Distribute [tex]\(-6y^2\)[/tex]:
[tex]\[ -6y^2 \cdot y^2 + (-6y^2) \cdot 2 \][/tex]
This gives us:
[tex]\[ -6y^4 - 12y^2 \][/tex]
3. Combine all the terms:
[tex]\[ 3y^2 + 6 - 6y^4 - 12y^2 \][/tex]
4. Combine like terms:
[tex]\[ -6y^4 + 3y^2 - 12y^2 + 6 \][/tex]
[tex]\[ -6y^4 - 9y^2 + 6 \][/tex]
### Conclusion from Clara's Work
Let's compare Clara's work with the correct method:
- Clara started with:
[tex]\[ \left(3 - 6y^2\right)\left(y^2 + 2\right) = 3 \left(y^2\right) + \left(-6y^2\right)(2) \][/tex]
- According to Clara, this resulted in:
[tex]\[ 3y^2 - 12y^2 \][/tex]
- And concluded:
[tex]\[ -9y^2 \][/tex]
### Identified Errors in Clara's Work
1. Clara missed the distributive steps involving [tex]\( -6y^2 \cdot y^2 \)[/tex].
2. Clara miscalculated the second distribution: while [tex]\(-6y^2\)[/tex] by [tex]\(2\)[/tex] was correctly calculated, the first term [tex]\(-6y^2 \cdot y^2\)[/tex] was ignored.
### Verifying Statements
Given the analysis above:
- "No, she did not multiply [tex]\(-6y^2\)[/tex] by 2 correctly." — This statement is incorrect because she actually multiplied [tex]\(-6y^2\)[/tex] by 2 correctly as [tex]\(-12y^2\)[/tex].
- "No, she did not add [tex]\(3y^2\)[/tex] and [tex]\(-12y^2\)[/tex] correctly." — This statement is incorrect because she subtracted correctly to get [tex]\(-9y^2\)[/tex].
- "No, she did not use the distributive property correctly." — This statement is correct because she did not distribute [tex]\( -6y^2 \cdot y^2\)[/tex] correctly.
- "Yes, she multiplied the binomials correctly." — This statement is incorrect based on our analysis.
### Final Answer
The correct statement is:
- No, she did not use the distributive property correctly.
### Correct Method Using the Distributive Property
To multiply the binomials, we should use the distributive property to distribute each term inside the first parenthesis to each term inside the second parenthesis. The distributive property states:
[tex]\[ (a + b)(c + d) = ac + ad + bc + bd \][/tex]
For our given binomials, [tex]\((3 - 6y^2)\)[/tex] and [tex]\((y^2 + 2)\)[/tex], we perform the following steps:
1. Distribute [tex]\(3\)[/tex]:
[tex]\[ 3 \cdot y^2 + 3 \cdot 2 \][/tex]
This gives us:
[tex]\[ 3y^2 + 6 \][/tex]
2. Distribute [tex]\(-6y^2\)[/tex]:
[tex]\[ -6y^2 \cdot y^2 + (-6y^2) \cdot 2 \][/tex]
This gives us:
[tex]\[ -6y^4 - 12y^2 \][/tex]
3. Combine all the terms:
[tex]\[ 3y^2 + 6 - 6y^4 - 12y^2 \][/tex]
4. Combine like terms:
[tex]\[ -6y^4 + 3y^2 - 12y^2 + 6 \][/tex]
[tex]\[ -6y^4 - 9y^2 + 6 \][/tex]
### Conclusion from Clara's Work
Let's compare Clara's work with the correct method:
- Clara started with:
[tex]\[ \left(3 - 6y^2\right)\left(y^2 + 2\right) = 3 \left(y^2\right) + \left(-6y^2\right)(2) \][/tex]
- According to Clara, this resulted in:
[tex]\[ 3y^2 - 12y^2 \][/tex]
- And concluded:
[tex]\[ -9y^2 \][/tex]
### Identified Errors in Clara's Work
1. Clara missed the distributive steps involving [tex]\( -6y^2 \cdot y^2 \)[/tex].
2. Clara miscalculated the second distribution: while [tex]\(-6y^2\)[/tex] by [tex]\(2\)[/tex] was correctly calculated, the first term [tex]\(-6y^2 \cdot y^2\)[/tex] was ignored.
### Verifying Statements
Given the analysis above:
- "No, she did not multiply [tex]\(-6y^2\)[/tex] by 2 correctly." — This statement is incorrect because she actually multiplied [tex]\(-6y^2\)[/tex] by 2 correctly as [tex]\(-12y^2\)[/tex].
- "No, she did not add [tex]\(3y^2\)[/tex] and [tex]\(-12y^2\)[/tex] correctly." — This statement is incorrect because she subtracted correctly to get [tex]\(-9y^2\)[/tex].
- "No, she did not use the distributive property correctly." — This statement is correct because she did not distribute [tex]\( -6y^2 \cdot y^2\)[/tex] correctly.
- "Yes, she multiplied the binomials correctly." — This statement is incorrect based on our analysis.
### Final Answer
The correct statement is:
- No, she did not use the distributive property correctly.
