Question

Factor the polynomial p(x) by grouping and write it in complete factored form.
p(x) = 4x5 – 32x2 – x
3 + 8
A) (2x + 1) (2x + 1) (x – 2) (x2 – 2x + 4)
B) (2x + 1) (2x – 1) (x + 2) (x2 + 2x + 4)
C) (2x – 1) (2x – 1) (x – 2)(x2 – 2x + 4)
D) (2x – 1) (2x + 1) (x – 2) (x2 + 2x + 4)

Answer 1

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:

1. From the first group \( 4x^5 - x^3 \), we can factor out \( x^3 \):

\[
x^3(4x^2 - 1)
\]

2. 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:

\[
\text{D) } (2x - 1)(2x + 1)(x - 2)(x^2 + 2x + 4)
\]