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 \( y^4 - 8y^2 + 16 \), 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**: \((a - b)^2 = a^2 - 2ab + b^2\).