To factor the polynomial \( y^4 - 8y^2 + 16 \), we can recognize that it can be rewritten in a form that matches one of the patterns provided.
First, observe that the polynomial can be treated as a quadratic in terms of \( y^2 \):
Let \( x = y^2 \), then the polynomial becomes: \[ x^2 - 8x + 16 \]
Now, we can factor \( x^2 - 8x + 16 \) using Pattern #2, which states that: \[ (a - b)^2 = a^2 - 2ab + b^2 \] Here, if we set \( a = x \) and \( b = 4 \), we can see that: \[ (x - 4)^2 = x^2 - 2(4)x + 16 \]
Thus, we can factor as follows: \[ x^2 - 8x + 16 = (x - 4)^2 \]
Substituting back \( x = y^2 \), we have: \[ y^4 - 8y^2 + 16 = (y^2 - 4)^2 \]
Now we can apply Pattern #3 to factor \( y^2 - 4 \): \[ y^2 - 4 = (y - 2)(y + 2) \]
Therefore, the complete factorization of the original polynomial is: \[ y^4 - 8y^2 + 16 = (y^2 - 4)^2 = ((y - 2)(y + 2))^2 \]
In conclusion, the pattern used to factor the polynomial is Pattern #2: \((a - b)^2 = a^2 - 2ab + b^2\).