Answer :
Certainly! Let's go through Cecile's work step-by-step to see how she factored the polynomial [tex]\(16x^6 - 9\)[/tex] using the X method.
1. Initial Expression:
The original expression is [tex]\(16x^6 - 9\)[/tex]. Cecile rewrites this as [tex]\(16x^6 + 0x - 9\)[/tex]. Adding [tex]\(0x\)[/tex] does not change the value of the polynomial and is often a step used to maintain structure for factoring.
2. Factoring Setup:
Since this expression is a difference of squares, Cecile goes on to use the appropriate identity: [tex]\((a^2 - b^2) = (a + b)(a - b)\)[/tex]. Here, [tex]\(16x^6\)[/tex] can be seen as [tex]\((4x^3)^2\)[/tex] and [tex]\(9\)[/tex] as [tex]\(3^2\)[/tex].
3. Decomposition:
She breaks down the expression in step 3 to emphasize the grouping needed:
[tex]\[
16x^6 + 12x^3 - 12x^3 - 9
\][/tex]
While it seems unnecessary in this context since no middle terms actually exist, it's part of the factoring strategy using pairs.
4. Factoring by Grouping:
Cecile factors by grouping:
[tex]\[
4x^3(4x^3 + 3) + (-3)(4x^3 + 3)
\][/tex]
She successfully identifies [tex]\((4x^3 + 3)\)[/tex] as a common factor.
5. Final Factored Form:
She combines the grouped factors:
[tex]\[
(4x^3 + 3)(4x^3 - 3)
\][/tex]
This is indeed correctly factored as it matches the difference of squares formula we observed in step 2.
Cecile's work is correct. She has correctly identified and factored the expression [tex]\(16x^6 - 9\)[/tex] into [tex]\((4x^3 + 3)(4x^3 - 3)\)[/tex].
1. Initial Expression:
The original expression is [tex]\(16x^6 - 9\)[/tex]. Cecile rewrites this as [tex]\(16x^6 + 0x - 9\)[/tex]. Adding [tex]\(0x\)[/tex] does not change the value of the polynomial and is often a step used to maintain structure for factoring.
2. Factoring Setup:
Since this expression is a difference of squares, Cecile goes on to use the appropriate identity: [tex]\((a^2 - b^2) = (a + b)(a - b)\)[/tex]. Here, [tex]\(16x^6\)[/tex] can be seen as [tex]\((4x^3)^2\)[/tex] and [tex]\(9\)[/tex] as [tex]\(3^2\)[/tex].
3. Decomposition:
She breaks down the expression in step 3 to emphasize the grouping needed:
[tex]\[
16x^6 + 12x^3 - 12x^3 - 9
\][/tex]
While it seems unnecessary in this context since no middle terms actually exist, it's part of the factoring strategy using pairs.
4. Factoring by Grouping:
Cecile factors by grouping:
[tex]\[
4x^3(4x^3 + 3) + (-3)(4x^3 + 3)
\][/tex]
She successfully identifies [tex]\((4x^3 + 3)\)[/tex] as a common factor.
5. Final Factored Form:
She combines the grouped factors:
[tex]\[
(4x^3 + 3)(4x^3 - 3)
\][/tex]
This is indeed correctly factored as it matches the difference of squares formula we observed in step 2.
Cecile's work is correct. She has correctly identified and factored the expression [tex]\(16x^6 - 9\)[/tex] into [tex]\((4x^3 + 3)(4x^3 - 3)\)[/tex].
