Answer:
(4x^2 + 1)(2x + 1)(2x - 1)
Step-by-step explanation:
16x^4 - 1
The first rule of factoring is to try to factor a common factor. There is no common factor between 16x^4 and 1.
We see that the polynomial is a binomial. It is a difference of two terms. We also notice that the first term is a perfect square. 16x^4 is the square of 4x^2. 1 is also a perfect square. It is the square of 1. Since both terms are perfect squares, and it is a difference, this binomial is the difference of two squares.
Recall the factoring of the difference of two squares. The difference of two squares factors into the product of a sum and a difference.
The pattern is:
a^2 - b^2 = (a + b)(a - b)
Apply that pattern to our problem. 16x^4 is the square of 4x^2, so the factorization looks like this:
16x^4 - 1 = (4x^2 + 1)(4x^2 - 1)
So far so good, but we are not finished yet. Factoring a polynomial means factoring it completely. Now we need to look at each factor and see if it can be factored further.
First, look at 4x^2 + 1. 4x^2 is the square of 2x. 1 is the square of 1. This is the sum of two squares. There is no factorization for the sum of two squares, so that factor is done.
Now look at 4x^2 - 1. Again, 4x^2 is the square of 2x, and 1 is the square of 1. In this case, we have the difference of two squares. We use the same factorization pattern we used above, a^2 b^2 = (a + b)(a - b).
16x^4 - 1 = (4x^2 + 1)(2x + 1)(2x - 1)
2x + 1 and 2x - 1 are not factorable, so the fully factored binomial is:
(4x^2 + 1)(2x + 1)(2x - 1)
