College

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][tex]$(4x^3 + 3)(4x^3 - 3)$[/tex][/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][tex]$16x^6 - 9$[/tex][/tex].
D. Yes, Cecile factored the polynomial correctly.

Answer :

Sure! Let's take a closer look at what Cecile did in her work to factor the polynomial [tex]\(16x^6 - 9\)[/tex].

1. Identifying the Type of Polynomial:
The expression [tex]\(16x^6 - 9\)[/tex] is a difference of squares. This is because it can be rewritten as [tex]\((4x^3)^2 - 3^2\)[/tex].

2. Applying the Difference of Squares Formula:
The difference of squares formula states that [tex]\(a^2 - b^2 = (a + b)(a - b)\)[/tex]. Here, [tex]\(a\)[/tex] is [tex]\(4x^3\)[/tex] and [tex]\(b\)[/tex] is [tex]\(3\)[/tex].

3. Factoring the Expression:
Using the formula:
[tex]\[
(4x^3)^2 - 3^2 = (4x^3 + 3)(4x^3 - 3)
\][/tex]

4. Reviewing Cecile's Steps:
- She wrote [tex]\(16x^6 + 0x - 9\)[/tex] to show that there is no linear term (the [tex]\(0x\)[/tex] is just a placeholder and doesn't affect the polynomial). This helps illustrate the process but isn't necessary in the factorization.
- She used the structure of the difference of squares to split the expression into [tex]\((4x^3 + 3)\)[/tex] and [tex]\((4x^3 - 3)\)[/tex].

5. Final Answer Verification:
Cecile's factorization is correct because the solution [tex]\((4x^3 + 3)(4x^3 - 3)\)[/tex] matches the expected result using the difference of squares method.

Thus, the correct analysis is: Yes, Cecile factored the polynomial correctly.