Question

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

Answer 1

\[\large 16x^{4}-1\]
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?

Answer 2

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
\[\large 4x^{2}-1=(?\ ?)(?\ ?)\]

Answer 3

2x-1 ?

Answer 4

The factors of
\[\large 4x^{2}-1=(2x+1)(2x-1)\]

Answer 5

Which is one of the answer choices.

Answer 6

ohh i see it now thanks~

Answer 7

You're welcome :)