Question

which pattern can be used to factor the high speed drgree 16x^8-49x^2
\left(a-b\right)^2=a^2-2ab+b^2
\left(a+b\right)^2=a^2+2ab+b^2
\left(a-b\right)\left(a+b\right)=a^2-b^2
16x^8-49x^2

Answer 1

To factor the expression \( 16x^8 - 49x^2 \), we can first notice that it can be treated as a difference of squares. The expression can be rewritten as follows:

1. Recognize the terms:
- \( 16x^8 \) can be rewritten as \( (4x^4)^2 \)
- \( 49x^2 \) can be rewritten as \( (7)^2 \)

2. Substitute these into the expression:
\[
16x^8 - 49x^2 = (4x^4)^2 - (7)^2
\]

3. Now, we apply the difference of squares formula \( a^2 - b^2 = (a-b)(a+b) \):
\[
= (4x^4 - 7)(4x^4 + 7)
\]

Thus, the factorization of \( 16x^8 - 49x^2 \) is:
\[
(4x^4 - 7)(4x^4 + 7)
\]

So, the pattern used here is \( (a-b)(a+b) = a^2 - b^2 \).