Help with factoring? I really dont get this stuff!!

I need a walk through on how to do these because im really not getting it!

n^2- 100

9k^2 + 4k^2+ 16

4x^2- 49

you dont have to solve all of them :) thanksss for help

The first and last one are called difference-of-squares. Both terms are squares and the second is subtracted from the first. These follow a formula:

a^2 - b^2 = (a + b)(a - b)

Substitute n for a and 10 (the square root of 100) for b to get:

n^2 - 100 = (n + 10)(n - 10)


I don't believe that the second problem will factor to integers, if it will factor at all. You can check the discriminant (the b^2 - 4ac term) of the quadratic equation. Use 9 for a, 4 for b, and 16 for c:

4^2 - 4(9)(16)

16 - 576 = -550

Since that's negative, a quadratic equation using that expression would have complex roots and it can't be factored to real numbers.

Numbers 1 & 3 are the difference of two squares;

(a + b)(a - b) = a² + ba - ab - b²
=a² - b²

So wherever we have a difference of two squares, we can write it as (a + b)(a - b)

So n² - 100 = (n + 10)(n - 10)
4x² - 49 = (2x + 7)(2x - 7)

For the 2nd one, are there really meant to be two "k²" terms?
If so, then it simplifies to 13k² + 16, and doesn't factor nicely.

you need to think of what to numbers multiplied will give you this bracket
since its n^2 then there should be 2 brackets and since its two digits then one bracket is addition and the other is subtraction
1- (n-10)(n+10)
there is something wrong in the second one
3- (2x-7)(2x+7)

n^2 - 100; what number when multiplied by itself gives you 100?

(n - 10)(n + 10) <======Answer

*Note: They have to have opposite signs so the terms in the middle cancel


4x^2 - 49

(2x - 7)(2x + 7)


Blessings