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Factor completely [tex]$16 x^8 - 1$[/tex].

A. [tex](4x^4 - 1)(4x^4 + 1)[/tex]

B. [tex](2x^2 - 1)(2x^2 + 1)(4x^4 + 1)[/tex]

C. [tex](2x^2 - 1)(2x^2 + 1)(2x^2 + 1)(2x^2 + 1)[/tex]

D. [tex](2x^2 - 1)(2x^2 + 1)(4x^4 - 1)[/tex]

Sagot :

GinnyAnswer
To completely factor the polynomial [tex]\( 16x^8 - 1 \)[/tex], we can follow these steps:

1. Recognize the difference of squares:
[tex]\[ 16x^8 - 1 = (4x^4)^2 - 1^2 \][/tex]
The difference of squares formula [tex]\( a^2 - b^2 = (a - b)(a + b) \)[/tex] allows us to rewrite this as:
[tex]\[ (4x^4 - 1)(4x^4 + 1) \][/tex]

Let's further factor each part:

2. Factor the first part, [tex]\( 4x^4 - 1 \)[/tex], again as a difference of squares:
[tex]\[ 4x^4 - 1 = (2x^2)^2 - 1^2 = (2x^2 - 1)(2x^2 + 1) \][/tex]

3. Check if the second part, [tex]\( 4x^4 + 1 \)[/tex], can be factored further:
The term [tex]\( 4x^4 + 1 \)[/tex] can further be examined using advanced factorization techniques, breaking it down:
[tex]\[ 4x^4 + 1 = (2x^2)^2 + (1)^2 \][/tex]
This can be expressed using advanced factorization as follows:
[tex]\[ 4x^4 + 1 = (2x^2 - 2x + 1)(2x^2 + 2x + 1) \][/tex]

4. Combine all the factors:
Putting it all together, we get:
[tex]\[ 16x^8 - 1 = (2x^2 - 1)(2x^2 + 1)(2x^2 - 2x + 1)(2x^2 + 2x + 1) \][/tex]

Therefore, the completely factored form of [tex]\( 16x^8 - 1 \)[/tex] is:
[tex]\[ (2x^2 - 1)(2x^2 + 1)(2x^2 - 2x + 1)(2x^2 + 2x + 1) \][/tex]
wegnerkolmp2741o

Answer:

B. (2x^2 - 1)(2x^2 + 1)(4x^4 + 1)

Step-by-step explanation:

Factor completely 16 x^8 - 1.

This is the difference of squares, a^2 - b^2 = (a-b) (a+b)

(4x^4)^2 - 1^2  where 4x^4 = a  and 1 = b

16x^8 -1 =  (4x^4 -1) (4x^4+1)

4x^4 -1 is also a difference of squares where a = 2x^2 and b=1

16x^8 -1 =  (2x^2 -1)(2x^2+1) (4x^4+1)

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