To factor the polynomial \( p(x) = 4x^5 - 32x^2 - x^3 + 8 \) by grouping, we will first rearrange the terms for easier grouping:
\[ p(x) = 4x^5 - x^3 - 32x^2 + 8 \]
Now we will group the terms:
\[ p(x) = (4x^5 - x^3) + (-32x^2 + 8) \]
Next, we can factor each group:
- From the first group \( 4x^5 - x^3 \), we can factor out \( x^3 \):
\[ x^3(4x^2 - 1) \]
- From the second group \( -32x^2 + 8 \), we can factor out \(-8\):
\[ -8(4x^2 - 1) \]
Putting the factored groups together, we have:
\[ p(x) = x^3(4x^2 - 1) - 8(4x^2 - 1) \]
Notice that both terms have a common factor of \( (4x^2 - 1) \):
\[ p(x) = (4x^2 - 1)(x^3 - 8) \]
Next, we can factor \( 4x^2 - 1 \) as a difference of squares:
\[ 4x^2 - 1 = (2x - 1)(2x + 1) \]
Also, \( x^3 - 8 \) is a difference of cubes, which can be factored using the formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \):
\[ x^3 - 8 = (x - 2)(x^2 + 2x + 4) \]
Now we can write the complete factored form of \( p(x) \):
\[ p(x) = (2x - 1)(2x + 1)(x - 2)(x^2 + 2x + 4) \]
This matches choice \( D \):
\[ D) (2x - 1)(2x + 1)(x - 2)(x^2 + 2x + 4) \]
Thus, the correct answer is:
\[ \text{D) } (2x - 1)(2x + 1)(x - 2)(x^2 + 2x + 4) \]
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