anonymous
  • anonymous
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
Mathematics
schrodinger
  • schrodinger
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kropot72
  • kropot72
\[\large 16x^{4}-1\] 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?
kropot72
  • kropot72
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 \[\large 4x^{2}-1=(?\ ?)(?\ ?)\]
anonymous
  • anonymous
2x-1 ?

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kropot72
  • kropot72
The factors of \[\large 4x^{2}-1=(2x+1)(2x-1)\]
kropot72
  • kropot72
Which is one of the answer choices.
anonymous
  • anonymous
ohh i see it now thanks~
kropot72
  • kropot72
You're welcome :)

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