To factor the polynomial \( y^4 - 8y^2 + 16 \), let's first identify if it can be rewritten in a form that matches any of the given patterns.
Notice that \( y^4 - 8y^2 + 16 \) can be treated as a quadratic in terms of \( y^2 \). Let's set \( x = y^2 \). Then, the polynomial becomes:
\[ x^2 - 8x + 16 \]
Now, we can try to factor this quadratic expression. We can see if it fits the pattern of a perfect square trinomial, which is described by Pattern #2: \( (a−b)^2 = a^2−2ab+b^2 \).
When we attempt to factor \( x^2 - 8x + 16 \), we notice:
\[ x^2 - 8x + 16 = (x - 4)^2 \]
Substituting back \( x = y^2 \):
\[ (y^2 - 4)^2 \]
Now, we can apply Pattern #3: \( (a−b)(a+b) = a^2−b^2 \) to factor \( y^2 - 4 \):
\[ y^2 - 4 = (y - 2)(y + 2) \]
Putting it all together, we have:
\[ y^4 - 8y^2 + 16 = (y^2 - 4)^2 = (y - 2)^2 (y + 2)^2 \]
In conclusion, the pattern used to factor \( y^4 - 8y^2 + 16 \) is a combination of:
- Recognizing a quadratic form (not a specific pattern).
- Using Pattern #2 to determine it is a perfect square.
- Finally applying Pattern #3 to further factor it.
Thus, the factorization is:
\[ (y - 2)^2(y + 2)^2 \]
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