Question

Factor

25m^100 − 121n^16

1. (5m^50 - 11n^8)^2
2. (5m^10 - 11n^4) (5m^10 + 11n^4)
3. (5m^50 - 11n^8) (5m^50 + 11n^8)
4. (5m^10 - 11n^4)^2

Answer 1

The correct answer is C) (5m^50 - 11n^8) (5m^50 + 11n^8)

We can tell this because of the rule regarding factoring the difference of two perfect squares. When we have two squares being multiplied, we can use the following rule.

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

In this case, or first term is 25m^100. So we can solve that by setting it equal to a^2.

a^2 = 25m^100 -----> take the square root of both sides

a = 5m^50

Then we can do the same for the b term.

b^2 = 121n^16 ----->take the square root of both sides

b = 11n^8

Now we can use both in the equation already given

(a - b)(a + b)

(5m^50 - 11n^16)(5m^50 + 11n^16)