Question

Factor the following high-degree polynomial completely.
48x^6-3x^2





A
3x2(4x2+1)(4x+1)(4x−1)3x^2\left(4x^2+1\right)\left(4x+1\right)\left(4x-1\right)3x
2
(4x
2
+1)(4x+1)(4x−1)

B
3x2(4x2+1)(4x2−1)3x^2\left(4x^2+1\right)\left(4x^2-1\right)3x
2
(4x
2
+1)(4x
2
−1)

C
3x2(8x2+1)(8x2−1)3x^2\left(8x^2+1\right)\left(8x^2-1\right)3x
2
(8x
2
+1)(8x
2
−1)

D
3x2(4x2+1)(2x+1)(2x−1)3x^2\left(4x^2+1\right)\left(2x+1\right)\left(2x-1\right)3x
2
(4x
2
+1)(2x+1)(2x−1)

Answer 1

To factor the polynomial \( 48x^6 - 3x^2 \) completely, we start with common factor extraction:

1. **Factor out the greatest common factor (GCF)**:
\[
48x^6 - 3x^2 = 3x^2(16x^4 - 1)
\]

2. **Factor the difference of squares**:
The term \(16x^4 - 1\) is a difference of squares:
\[
16x^4 - 1 = (4x^2)^2 - (1)^2 = (4x^2 - 1)(4x^2 + 1)
\]

3. **Further factor \(4x^2 - 1\)**:
The term \(4x^2 - 1\) can also be factored as a difference of squares:
\[
4x^2 - 1 = (2x)^2 - (1)^2 = (2x - 1)(2x + 1)
\]

4. **Putting it all together**:
Now we can rewrite the factored form:
\[
48x^6 - 3x^2 = 3x^2 (4x^2 + 1) (2x - 1)(2x + 1)
\]

Thus, the complete factorization of the polynomial is:
\[
3x^2 (4x^2 + 1)(2x - 1)(2x + 1)
\]

### Conclusion:
Comparing this with the options provided:
- **Option D** is correct: \( 3x^2(4x^2 + 1)(2x + 1)(2x - 1) \) matches our factorization.

So, the answer is **D**.