High School

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 :

We begin with the polynomial

[tex]$$16x^6 - 9.$$[/tex]

Notice that it can be written as a difference of perfect squares:

[tex]$$16x^6 - 9 = (4x^3)^2 - 3^2.$$[/tex]

Using the difference of squares formula

[tex]$$a^2 - b^2 = (a + b)(a - b),$$[/tex]

with [tex]$a = 4x^3$[/tex] and [tex]$b = 3$[/tex], we factor the expression as

[tex]$$ (4x^3 + 3)(4x^3 - 3). $$[/tex]

In Cecile's work, she rewrites the polynomial as

[tex]$$16x^6 + 12x^3 - 12x^3 - 9,$$[/tex]

which is completely equivalent because the terms [tex]$+12x^3$[/tex] and [tex]$-12x^3$[/tex] cancel each other out. This allows her to use the grouping method (often called the "X method"). She groups the terms as follows:

[tex]$$4x^3(4x^3 + 3) - 3(4x^3 + 3)$$[/tex]

and then factors out the common binomial [tex]$(4x^3 + 3)$[/tex]:

[tex]$$ (4x^3 + 3)(4x^3 - 3). $$[/tex]

Thus, Cecile's factorization is correct.

The final factors of the polynomial are

[tex]$$4x^3 + 3 \quad \text{and} \quad 4x^3 - 3.$$[/tex]

Therefore, the correct answer is that Cecile factored the polynomial correctly.