Factorise x to the power of 4n + x to the power of 2n + 1 completely where n is an odd integer.​

Question

Factorise x to the power of 4n + x to the power of 2n + 1 completely where n is an odd integer.​

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Sigridomena 3 years 2021-08-26T01:16:11+00:00 2 Answers 9 views 0

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  1. Latifah
    0
    2021-08-26T01:17:57+00:00
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    ANSWER:

    Let x = (−1)^n +(−1)^2n +(−1)^2n+1 +(−1)^4n + 1 where n is odd.

    (−1)^n = − 1

    (−1)^2n = 1

    (−1)^2n + 1 = (−1)

    2n(−1) = − 1

    (−1)^4n + 1 = (−1)

    4n(−1) = − 1

    ∴ x= − 1 + 1 − 1 − 1 = −2

  2. thongdat2
    0
    2021-08-26T01:17:57+00:00
    Reply

    Answer:

    • x^{3}x^{4m}(x^{4m+1} + 1)

    ———————————————

    • x⁴ⁿ + x²ⁿ⁺¹ = x²ⁿ⁺¹(x²ⁿ⁻¹ + 1)

    n is an odd integer, let n = 2m + 1

    Then

    • 2n – 1 = 2(2m + 1) – 1 = 4m + 2 – 1 = 4m + 1
    • 2n + 1 = 2(2m + 1) + 1 = 4m + 2 + 1 = 4m + 3
    • x²ⁿ⁺¹(x²ⁿ⁻¹ + 1) =
    • x^{4m + 3}(x^{4m+1} + 1) =
    • x^{3}x^{4m}(x^{4m+1} + 1)

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