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Math Help - Factorisation

  1. #1
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    Question Factorisation

    (1 + x + x^2 + x^3)^2 - x^3
    Please help me .....
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  2. #2
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    Hello anshulbshah
    Quote Originally Posted by anshulbshah View Post
    (1 + x + x^2 + x^3)^2 - x^3
    Please help me .....
    See the attached graph, which shows that f(x)=(1 + x + x^2 + x^3)^2 - x^3 =0 doesn't have any real roots, and hence (1 + x + x^2 + x^3)^2 - x^3 doesn't have any factors.

    f(-1)=f(0)=1, and there's a minimum turning point between these two values, but this minimum value is greater than zero. You could possibly prove this by calculus.

    For x>0 and x<-1, f(x) increases very rapidly, since the dominant term when the brackets are expanded is x^6.

    Grandad
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  3. #3
    Senior Member pacman's Avatar
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    it can be solve by grouping after expanding,

    (1 + x + x^2 + x^3)^2 - x^3 = (x^2 + x + 1)(x^4 + x^3 + x^2 + x + 1)
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  4. #4
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    Quote Originally Posted by pacman View Post
    it can be solve by grouping after expanding,

    (1 + x + x^2 + x^3)^2 - x^3 = (x^2 + x + 1)(x^4 + x^3 + x^2 + x + 1)
    Thanks for this - I was thinking of a linear factor, of course!

    Grandad
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  5. #5
    Senior Member pacman's Avatar
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    Prove: (1 + x + x^2 + x^3)^2 - x^3 = (x^2 + x + 1)(x^4 + x^3 + x^2 + x + 1)

    [(1 + x) + (x^2 + x^3)]^2 - x^3 = (1 + x)^2 + 2(1 + x)(x^2 + x^3) + (x^2 + x^3)^2 - x^3,

    = (1 + 2x + x^2) + (2x^2 + 2x^3 +2x^3 + 2x^4) + (x^4 + 2x^5 + x^6) - x^3, rearranging

    = (1) + (2x) + (x^2 + 2x^2) + (2x^3 +2x^3 - x^3) + (2x^4 + x^4) + (2x^5) + (x^6),

    = 1 + 2x + 3x^2 + 3x^3 + 3x^4 + 2x^5 + x^6, sub-divide with numerical coefficient equal to 1, group them THEN apply factoring

    = (1 + x + x^2) + (x + x^2 + x^3) + (x^2 + x^3 + x^4) + (x^3 + x^4 + x^5) + (x^4 + x^5 + x^6)

    = (1 + x + x^2) + x(1 + x + x^2) + x^2(1 + x + x^2) + x^3(1 + x + x^2) + x^4(1 + x + x^2), factoring

    = (1 + x + x^2)(1 + x + x^2 + x^3 + x^4).






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