Question

which pattern can be used to factor the high-degree polynomial 16x^8−49x^2 ?

Answer 1

To factor the polynomial \( 16x^8 - 49x^2 \), we can start by looking for a common factor and then apply the difference of squares pattern.

1. **Factor out the common term:** Both terms have a common factor of \( x^2 \):

\[
16x^8 - 49x^2 = x^2(16x^6 - 49)
\]

2. **Recognize the difference of squares:** The expression \( 16x^6 - 49 \) can be recognized as a difference of squares. Recall that \( a^2 - b^2 = (a - b)(a + b) \).

Here, we can identify:
- \( a = 4x^3 \) (since \( (4x^3)^2 = 16x^6 \))
- \( b = 7 \) (since \( 7^2 = 49 \))

Thus, we can rewrite \( 16x^6 - 49 \):

\[
16x^6 - 49 = (4x^3 - 7)(4x^3 + 7)
\]

3. **Combine everything:** Now substitute back to get the complete factorization:

\[
16x^8 - 49x^2 = x^2(4x^3 - 7)(4x^3 + 7)
\]

So, the fully factored form of the polynomial \( 16x^8 - 49x^2 \) is:

\[
x^2(4x^3 - 7)(4x^3 + 7)
\]