Answer :
Let's analyze Cecile's work to see if she factored the polynomial [tex]\(16x^6 - 9\)[/tex] correctly using the X method.
### Step 1: Understanding the Polynomial
The original polynomial is [tex]\(16x^6 - 9\)[/tex]. This is a difference of squares because it can be expressed as:
[tex]\[ (4x^3)^2 - 3^2 \][/tex]
### Step 2: Applying the Difference of Squares Formula
The difference of squares formula is:
[tex]\[ a^2 - b^2 = (a - b)(a + b) \][/tex]
For our polynomial:
- [tex]\(a = 4x^3\)[/tex]
- [tex]\(b = 3\)[/tex]
Applying the formula:
[tex]\[ (4x^3)^2 - 3^2 = (4x^3 - 3)(4x^3 + 3) \][/tex]
### Step 3: Examining Cecile’s Process
1. Cecile started with [tex]\(16x^6 + 0x - 9\)[/tex]. Introducing the [tex]\(0x\)[/tex] term does not affect the polynomial, so this step does not change its value.
2. In the X method, she decomposed the middle term, but since there is no middle term here initially, she split the expression directly into:
[tex]\[ 16x^6 + 12x^3 - 12x^3 - 9 \][/tex]
which simplifies back to [tex]\(16x^6 - 9\)[/tex].
3. Next, she factored by grouping:
[tex]\[ 4x^3(4x^3 + 3) + (-3)(4x^3 + 3) \][/tex]
4. This gives us the factors:
[tex]\[ (4x^3 + 3)(4x^3 - 3) \][/tex]
### Step 4: Conclusion
- Cecile's final factorization of [tex]\((4x^3 + 3)(4x^3 - 3)\)[/tex] is indeed the correct factorization of [tex]\(16x^6 - 9\)[/tex] using the difference of squares.
Therefore, the statement "Yes, Cecile factored the polynomial correctly" is accurate. Cecile’s work shows a correct application of the factorization of the difference of squares.
### Step 1: Understanding the Polynomial
The original polynomial is [tex]\(16x^6 - 9\)[/tex]. This is a difference of squares because it can be expressed as:
[tex]\[ (4x^3)^2 - 3^2 \][/tex]
### Step 2: Applying the Difference of Squares Formula
The difference of squares formula is:
[tex]\[ a^2 - b^2 = (a - b)(a + b) \][/tex]
For our polynomial:
- [tex]\(a = 4x^3\)[/tex]
- [tex]\(b = 3\)[/tex]
Applying the formula:
[tex]\[ (4x^3)^2 - 3^2 = (4x^3 - 3)(4x^3 + 3) \][/tex]
### Step 3: Examining Cecile’s Process
1. Cecile started with [tex]\(16x^6 + 0x - 9\)[/tex]. Introducing the [tex]\(0x\)[/tex] term does not affect the polynomial, so this step does not change its value.
2. In the X method, she decomposed the middle term, but since there is no middle term here initially, she split the expression directly into:
[tex]\[ 16x^6 + 12x^3 - 12x^3 - 9 \][/tex]
which simplifies back to [tex]\(16x^6 - 9\)[/tex].
3. Next, she factored by grouping:
[tex]\[ 4x^3(4x^3 + 3) + (-3)(4x^3 + 3) \][/tex]
4. This gives us the factors:
[tex]\[ (4x^3 + 3)(4x^3 - 3) \][/tex]
### Step 4: Conclusion
- Cecile's final factorization of [tex]\((4x^3 + 3)(4x^3 - 3)\)[/tex] is indeed the correct factorization of [tex]\(16x^6 - 9\)[/tex] using the difference of squares.
Therefore, the statement "Yes, Cecile factored the polynomial correctly" is accurate. Cecile’s work shows a correct application of the factorization of the difference of squares.
