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:

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