Recognize that the given expression is a perfect square trinomial in the form \(a^2-2ab+b^2\), where \(a=x^2\) and \(b=4\).
$$ a^2-2ab+b^2 = (a-b)(a-b)$$
$$ x^4-8x^2+16= (x^2)^2-2(x^2)(4)+(4)^2 $$
$$= (x^2-4)(x^2-4) $$
\(x^2-4\) is also a special quadratic, which can be factored using the difference of squares.
$$ a^2-b^2 = (a+b)(a-b) $$
$$ (x)^2-(2)^2= (x-2)(x+2) $$
To put it all together,
$$ x^4-8x^2+16 = (x^2-4)(x^2-4) $$
$$= (x+2)(x-2)(x+2)(x-2)$$
$$ =\boxed{(x+2)^2(x-2)^2}$$
If you have difficulty factoring, we can always expand the answer choices. Because the overall power of our expression is 4 \((x^4)\), we can eliminate the
last two answer choices, which both have an overall power of 2. It would take quite a while to show all the work for the expansion of \((x-2)^4\) and \( (x+2)^2 (x-2)^2\), but it may work for other questions.
Regardless, we can knock two answers out by simply looking at the overall power of the expression.
For equivalent expression problems, we can substitute a value for \(x\) and check it it satisfies the expressions that are supposed to be equal. Using \(x=1\),
$$ x^4-8x^2+16 $$
$$ (1)^4-8(1)^2+16 = 1-8+16=9$$
Choice 1
$$ (x-2)^4$$
$$ (1-2)^4 = (-1)^4 = 1 ✖ $$
Choice 2
$$ (x+2)^2 (x-2)^2 $$
$$ (1+2)^2 (1-2)^2 = (3)^2 (-1)^2 = 9 ✓ $$
Note: Usually we would use \(x=0\), but for this question all three expressions would be equal.