Show that 4n^3 + 6n^2 + 4n + 1 is a composite integer for each n E N.?

n E N = > n is an element of the set of natural numbers

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  • 1 decade ago
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    The following algebra uses binomial expansion as well as difference of two squares:

    4n^3+6n^2+4n+1 =

    4n^3+6n^2+4n+1+n^4-n^4 =

    n^4+4n^3+6n^2+4n+1-n^4 =

    (n+1)^4-n^4 =

    [ (n+1)^2-n^2 ] * [ (n+1)^2+n^2 ] = (2n+1)(2n^2+2n+1).

    So 4n^3+6n^2+4n+1 is always divisible by 2n+1. For any natural number n, 1 < 2n+1 < 4n^3+6n^2+4n+1, so 4n^3+6n^2+4n+1 is composite.

  • 1 decade ago

    4n^3 + 6n^2 + 4n + 1 = (2n+1)(2n^2 + 2n + 1). Since n is a natural number, both factors are bigger than 1, so it is composite.

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