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20 December, 07:51

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

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  1. 20 December, 11:19
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    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)
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