Question

which pattern can be used to factor the high speed drgree 16x^8-49x^2

Answer 1

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

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

This takes the shape of \( a^2 - b^2 \), where \( a = 4x^4 \) and \( b = 7 \). The difference of squares can be factored using the identity \( a^2 - b^2 = (a - b)(a + b) \):

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

Next, we can also factor the expression further if we factor out the greatest common factor from the original expression before applying the difference of squares:

1. First, notice that both terms share a common factor of \( x^2 \):

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

2. Now, we can factor \( 16x^6 - 49 \) using the same method, applying the difference of squares again:

\[
16x^6 - 49 = (4x^3)^2 - (7)^2
\]

This gives us another difference of squares:

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

Putting it all together, the fully factored expression is:

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

Thus, the final factored form of the expression \( 16x^8 - 49x^2 \) is:

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