Can someone explain this kind of ALGERBRA problem to me?..please?

Question

factoring polynomials?

16-49n^2

How do I factor this since there isnt a common number that goes into both 49 and 16?

I dont understand can someone explain this process on how to do this to me?

Answer 1

16-49n^2

(4+7n)(4-7n)

the way to check it is F.O.I.L (first, outer, inner, last)

4*4=16
4*7n=28n
4*-7n=-28n
7n*-7n=-49n^2

or 16+28n-28n-49n^2
28n and -28n cancel each other out and you are left with 16-49n^2

Answer 2

You don't need to find a common number. You are looking for expressions - that is letter number combinations.
In this case, use (4-7n) x (4+7n), which equals:
4 x4 +28n - 28n - 7 x 7 x n x n
Since the middle expression of the product is 28n - 28n, it cancels to zero, leaving only 16 - 49n^2

Answer 3

This is a difference of squares problem.
A^2 -B^2 = (A+B) x (A-B)