High School

Dear beloved readers, welcome to our website! We hope your visit here brings you valuable insights and meaningful inspiration. Thank you for taking the time to stop by and explore the content we've prepared for you.
------------------------------------------------ Cecile used the X method to factor [tex]16x^6 - 9[/tex].

1. [tex]16x^6 + 0x - 9[/tex]
2. X method:
3. [tex]16x^6 + 12x^3 - 12x^3 - 9[/tex]
4. [tex]4x^3(4x^3 + 3) + (-3)(4x^3 + 3)[/tex]
5. [tex](4x^3 + 3)(4x^3 - 3)[/tex]

Analyze Cecile's work. Is it correct?

A. No, adding in [tex]0x[/tex] keeps an equivalent polynomial.
B. No, she did not fill in the X correctly. She should have 16 on top and -9 on the bottom.
C. No, [tex]16x^6 + 12x^3 - 12x^3 - 9[/tex] is not equivalent to [tex]16x^6 - 9[/tex].
D. Yes, Cecile factored the polynomial correctly.

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].