Let's analyze the given expressions and determine which statement is correct.
Given expression: [tex]\((a^2 - 4)(a^4 + 4a^2 + 16)\)[/tex].
### 1. Equivalence Check
First, let's determine if the factored expression [tex]\((a^2 - 4)(a^4 + 4a^2 + 16)\)[/tex] is equivalent to the original expression [tex]\(a^6 - 64\)[/tex].
#### Multiplying the Factors
- [tex]\(a^2 - 4\)[/tex] can be written as [tex]\((a - 2)(a + 2)\)[/tex].
- We check [tex]\((a^4 + 4a^2 + 16)\)[/tex] by multiplying it with [tex]\((a^2 - 4)\)[/tex]:
[tex]\[
(a^2 - 4)(a^4 + 4a^2 + 16)
\][/tex]
Now, let’s expand this product:
[tex]\(
(a^2 - 4)(a^4 + 4a^2 + 16) = a^2(a^4 + 4a^2 + 16) - 4(a^4 + 4a^2 + 16)
\)[/tex]
[tex]\[
= a^6 + 4a^4 + 16a^2 - 4a^4 - 16a^2 - 64
\][/tex]
[tex]\[
= a^6 - 64
\][/tex]
So, the expression [tex]\((a^2 - 4)(a^4 + 4a^2 + 16)\)[/tex] is equivalent to [tex]\(a^6 - 64\)[/tex].
### 2. Complete Factoring Check
Now, let's check if each part of the expression [tex]\((a^2 - 4)(a^4 + 4a^2 + 16)\)[/tex] is completely factored.
#### Checking [tex]\(a^2 - 4\)[/tex]:
[tex]\[
a^2 - 4 = (a - 2)(a + 2)
\][/tex]
Since it can be further factored, [tex]\((a^2 - 4)\)[/tex] is not completely factored.
#### Checking [tex]\(a^4 + 4a^2 + 16\)[/tex]:
This expression can be thought of in terms of a substitution, let [tex]\(b = a^2\)[/tex]:
[tex]\[
b^2 + 4b + 16
\][/tex]
For a quadratic expression [tex]\(b^2 + 4b + 16\)[/tex], we need to check if it can be factored further. A quadratic expression can be factored if there are two numbers that multiply to the constant term (16) and add up to the linear coefficient (4). In this case, there are no such numbers, which means it cannot be factored further over the real numbers.
Thus, [tex]\(a^4 + 4a^2 + 16\)[/tex] is already completely factored.
### Conclusion
The expression [tex]\((a^2 - 4)(a^4 + 4a^2 + 16)\)[/tex] is equivalent to [tex]\(a^6 - 64\)[/tex], but the term [tex]\((a^2 - 4)\)[/tex] is not completely factored.
Therefore, the correct statement is:
The expression is equivalent, but the [tex]\(a^2 - 4\)[/tex] term is not completely factored.
