Question

Identify the pattern that can be used to factor the high-degree polynomial y4−8y2+16 . Pattern #1: (a+b)2=a2+2ab+b2 Pattern #2: (a−b)2=a2−2ab+b2 Pattern #3: (a−b)(a+b)=a2−b2 (1 point)

Answer 1

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:

1. Recognizing a quadratic form (not a specific pattern).
2. Using Pattern #2 to determine it is a perfect square.
3. Finally applying Pattern #3 to further factor it.

Thus, the factorization is:

\[
(y - 2)^2(y + 2)^2
\]