To factor the polynomial \(48x^6 - 3x^2\), we start by factoring out the greatest common factor (GCF).
The GCF of \(48x^6\) and \(-3x^2\) is \(3x^2\). Factoring that out, we have:
\[ 48x^6 - 3x^2 = 3x^2(16x^4 - 1) \]
Next, we notice that \(16x^4 - 1\) is a difference of squares:
\[ 16x^4 - 1 = (4x^2)^2 - 1^2 = (4x^2 - 1)(4x^2 + 1) \]
The term \(4x^2 - 1\) can be factored further, as it is also a difference of squares:
\[ 4x^2 - 1 = (2x - 1)(2x + 1) \]
Now, we can substitute everything back into our expression:
\[ 48x^6 - 3x^2 = 3x^2(4x^2 - 1)(4x^2 + 1) = 3x^2(2x - 1)(2x + 1)(4x^2 + 1) \]
So the complete factorization of the polynomial \(48x^6 - 3x^2\) is:
\[ 3x^2(2x - 1)(2x + 1)(4x^2 + 1) \]
However, the given options don't exactly match this factorization format. The closest option in the provided responses that matches our factorization is:
\[ 3x^2(4x^2 + 1)(2x + 1)(2x - 1) \]
b) \(3 x^2(4 x^2 + 1)(2 x + 1)(2 x - 1)\)