Answer :
We start with the polynomial
$$
16x^6 - 9.
$$
Notice that this expression can be written as a difference of two squares. Indeed, we can write
$$
16x^6 - 9 = (4x^3)^2 - 3^2.
$$
Since the difference of two squares factors as
$$
A^2 - B^2 = (A+B)(A-B),
$$
taking $A = 4x^3$ and $B = 3$, we immediately obtain
$$
16x^6 - 9 = (4x^3+3)(4x^3-3).
$$
Cecile's work begins by expressing the original polynomial as
$$
16x^6 + 0x - 9,
$$
which is equivalent to $16x^6-9$ because adding $0x$ does not change the value of the polynomial.
Then she rewrites $0x$ as $+12x^3-12x^3$, so that
$$
16x^6-9 = 16x^6 + 12x^3 - 12x^3 - 9.
$$
This step is valid because adding and subtracting the same term does not alter the polynomial.
Now, using factoring by grouping:
1. Group the first two terms and the last two terms:
$$
\left(16x^6+12x^3\right) + \left(-12x^3-9\right).
$$
2. Factor $4x^3$ from the first group:
$$
4x^3(4x^3+3),
$$
and factor $-3$ from the second group:
$$
-3(4x^3+3).
$$
3. Notice that both groups contain the common factor $(4x^3+3)$. Factoring it out gives:
$$
(4x^3+3)(4x^3-3).
$$
Thus, Cecile correctly factors the polynomial $16x^6 - 9$.
The final answer is: Yes, Cecile factored the polynomial correctly.
$$
16x^6 - 9.
$$
Notice that this expression can be written as a difference of two squares. Indeed, we can write
$$
16x^6 - 9 = (4x^3)^2 - 3^2.
$$
Since the difference of two squares factors as
$$
A^2 - B^2 = (A+B)(A-B),
$$
taking $A = 4x^3$ and $B = 3$, we immediately obtain
$$
16x^6 - 9 = (4x^3+3)(4x^3-3).
$$
Cecile's work begins by expressing the original polynomial as
$$
16x^6 + 0x - 9,
$$
which is equivalent to $16x^6-9$ because adding $0x$ does not change the value of the polynomial.
Then she rewrites $0x$ as $+12x^3-12x^3$, so that
$$
16x^6-9 = 16x^6 + 12x^3 - 12x^3 - 9.
$$
This step is valid because adding and subtracting the same term does not alter the polynomial.
Now, using factoring by grouping:
1. Group the first two terms and the last two terms:
$$
\left(16x^6+12x^3\right) + \left(-12x^3-9\right).
$$
2. Factor $4x^3$ from the first group:
$$
4x^3(4x^3+3),
$$
and factor $-3$ from the second group:
$$
-3(4x^3+3).
$$
3. Notice that both groups contain the common factor $(4x^3+3)$. Factoring it out gives:
$$
(4x^3+3)(4x^3-3).
$$
Thus, Cecile correctly factors the polynomial $16x^6 - 9$.
The final answer is: Yes, Cecile factored the polynomial correctly.
