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anonymous

  • one year ago

will give a medall!! one of the factors of 16x^4-1 is 4x^2+1. what are the other factors? 1)8x+1,x-1 2)4x+1,x-1 3)2x+1,2x-1 4)4x-1,x+1

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  1. kropot72
    • one year ago
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    \[\large 16x^{4}-1\] is the difference of two squares. The factorization of the difference of two squares is given by \[\large a^{2}-b^{2}=(a+b)(a-b)\] Can you now choose the correct answer?

  2. kropot72
    • one year ago
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    There are two steps to finding the solution. First step is to find the initial factors of the given expression as follows: \[\large 16x^{4}-1=(4x^{2}+1)(4x^{2}-1)\] It is given in the question that (4x^2 + 1) is one factor, therefore to find the other factors you need to factorize the perfect square \[\large 4x^{2}-1=(?\ ?)(?\ ?)\]

  3. anonymous
    • one year ago
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    2x-1 ?

  4. kropot72
    • one year ago
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    The factors of \[\large 4x^{2}-1=(2x+1)(2x-1)\]

  5. kropot72
    • one year ago
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    Which is one of the answer choices.

  6. anonymous
    • one year ago
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    ohh i see it now thanks~

  7. kropot72
    • one year ago
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    You're welcome :)

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