To factor the polynomial \( y^4 - 8y^2 + 16 \), we can start by looking for a substitution that simplifies the expression.
Notice that \( y^4 \) can be rewritten as \( (y^2)^2 \), and \( -8y^2 \) and \( +16 \) can be seen as a quadratic expression in \( y^2 \).
Let's use \( x = y^2 \). Then our polynomial becomes:
\[ x^2 - 8x + 16 \]
Next, we can identify that this resembles a perfect square trinomial. The structure \( x^2 - 8x + 16 \) follows the format of \( a^2 - 2ab + b^2 = (a - b)^2 \).
Here, we can recognize that:
- \( a = x \) (which is \( y^2 \))
- \( b = 4 \) (since \( -2ab = -8x \) implies \( b = \frac{8}{2a} = 4 \))
Thus, we have:
\[ x^2 - 8x + 16 = (x - 4)^2 \]
Substituting back \( x = y^2 \), we get:
\[ (y^2 - 4)^2 \]
Now, we can see if \( y^2 - 4 \) can be factored further. Recognizing \( y^2 - 4 \) as a difference of squares, we can apply Pattern #3:
\[ y^2 - 4 = (y - 2)(y + 2) \]
Putting this together, we find:
\[ (y^2 - 4)^2 = [(y - 2)(y + 2)]^2 = (y - 2)^2(y + 2)^2 \]
Thus, the complete factorization of the polynomial \( y^4 - 8y^2 + 16 \) is:
\[ (y - 2)^2 (y + 2)^2 \]
The appropriate pattern used is actually a combination of Pattern #2 (for the perfect square trinomial) and Pattern #3 (for the difference of squares).
1 answer