Factor. 25m100−121n16 (5m50−11n8)(5m50+11n8) (5m50−11n8)2 (5m10−11n4)2 (5m10−11n4)(5m10+11n4)

The answer to your question is (5m50−11n8)(5m50+11n8)

Step-by-step explanation:

                                             25m¹⁰⁰−121n¹⁶

find the prime factors of 25 and 100

    25     5                                  121     11

       5     5                                   11      11

        1                                            1

25 = 5 x 5 = 5²                       121 = 11 x 11 = 11²

m¹⁰⁰ = (m⁵⁰)²                           n¹⁶ = (n⁸)²

Substitution

                     25m¹⁰⁰−121n¹⁶ = 5²(m⁵⁰)²   - 11² (n⁸)²

                                            = (5m⁵⁰  - 11n⁸)  (5m⁵⁰  + 11n⁸)

(5m50−11n8)(5m50+11n8)

(5m50−11n8)2

(5m10−11n4)2

(5m10−11n4)(5m10+11n4)

  n=   0.0000 - 0.3750 i  

  n=   0.0000 + 0.3750 i

step-by-step explanation:

one of these should be correct