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Math Help - show equation and find the factor

  1. #1
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    show equation and find the factor

    2a)given that f(x) =x^n-nx+n-1 for the integer n>1. by considering f(x) and f'(x), show that f(x)=(X-1)^2 X g (x) is true for all polynomial g(x) with interger coefficients.

    hence , or otherwise
    1) show that 3^2n - 8n -1 is divisiblke by 64 for all intergers n>1
    2) show that the equation X^4 -4x+3 = 0

    . question2



    Moderator,

    i am learning abt factor . example 4x^4-4x^3-9x^2+x+2 . find the factor , factorise it! ( without using trial , any other method provided?
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  2. #2
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    Quote Originally Posted by qweiop90 View Post
    2a)given that f(x) =x^n-nx+n-1 for the integer n>1. by considering f(x) and f'(x), show that f(x)=(X-1)^2 X g (x) is true for all polynomial g(x) with interger coefficients.

    [snip]
    The result you have stated is wrong. The correct statement would be something like:

    Show that f(x) = (x-1)^2 g(x) for some polynomial g(x) with integer coefficients.



    First note that x = 1 is a root of f(x) and so x - 1 is a factor of f(x). Then f(x) = (x - 1) \, h(x).

    Now note that x = 1 is also a root of f'(x). Then x - 1 is a factor of f'(x) = h(x) + (x - 1) h'(x).

    It follows that (x - 1) is a factor of h(x). Therefore h(x) = (x - 1) g(x) and the result follows.
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  3. #3
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    Quote Originally Posted by qweiop90 View Post
    2a)given that f(x) =x^n-nx+n-1 for the integer n>1. by considering f(x) and f'(x), show that f(x)=(X-1)^2 X g (x) is true for all polynomial g(x) with interger coefficients.

    hence , or otherwise
    1) show that 3^2n - 8n -1 is divisiblke by 64 for all intergers n>1

    Mr F says: The result is simple to show using proof by induction.

    2) show that the equation X^4 -4x+3 = 0

    Mr F says: You probably mean

    Solve the equation x^4 - 4x + 3 = 0.

    Note that n = 4.

    [snip]
    ..
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  4. #4
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    Quote Originally Posted by qweiop90 View Post
    2a)given that f(x) =x^n-nx+n-1 for the integer n>1. by considering f(x) and f'(x), show that f(x)=(X-1)^2 X g (x) is true for all polynomial g(x) with interger coefficients.

    hence , or otherwise
    1) show that 3^2n - 8n -1 is divisiblke by 64 for all intergers n>1
    [snip]
    As an alternative to proof by induction, you can note that 3^{2n} = 9^n. So substitute x = 9 into the previous result:

    f(9) = (9-1)^2 g(9) = 64 g(9).

    Also f(9) = 9^n - 9n + n - 1 = 3^{2n} - 8n - 1.

    Therefore 64 g(9) = 3^{2n} - 8n - 1 ~ .....
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    The result you have stated is wrong. The correct statement would be something like:

    Show that f(x) = (x-1)^2 g(x) for some polynomial g(x) with integer coefficients.



    First note that x = 1 is a root of f(x) and so x - 1 is a factor of f(x). Then f(x) = (x - 1) \, h(x).

    Now note that x = 1 is also a root of f'(x). Then x - 1 is a factor of f'(x) = h(x) + (x - 1) h'(x).

    It follows that (x - 1) is a factor of h(x). Therefore h(x) = (x - 1) g(x) and the result follows.

    dont really understand? mind to show how the solution can be done?
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  6. #6
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    Quote Originally Posted by qweiop90 View Post
    dont really understand? mind to show how the solution can be done?
    What part don't you understand?
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    er show that . f(x) = (x-1)^2 g(x) . where is ^2 come frm? and how to prove?
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    Quote Originally Posted by qweiop90 View Post
    er show that . f(x) = (x-1)^2 g(x) . where is ^2 come frm? and how to prove?
    What part of my reply (post #2) don't you understand.
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    Now note that x = 1 is also a root of f'(x). Then x - 1 is a factor of .

    It follows that (x - 1) is a factor of h(x). Therefore and the result follows.

    . how it comes frm?
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  10. #10
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    Quote Originally Posted by qweiop90 View Post
    Now note that x = 1 is also a root of f'(x). Then x - 1 is a factor of .

    It follows that (x - 1) is a factor of h(x). Therefore and the result follows.

    . how it comes frm?
    Substitute into f(x) = (x - 1) \, h(x).
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  11. #11
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    First note that x = 1 is a root of f(x) and so x - 1 is a factor of f(x). Then .<---1

    Now note that x = 1 is also a root of f'(x). Then x - 1 is a factor of <---2


    why it become like this?
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  12. #12
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    Quote Originally Posted by qweiop90 View Post
    First note that x = 1 is a root of f(x) and so x - 1 is a factor of f(x). Then .<---1

    Now note that x = 1 is also a root of f'(x). Then x - 1 is a factor of <---2


    why it become like this?
    I got f(x) = (x-1) h(x) by noting that (x-1) is a factor of f(x). At this point I think it would be best if you went back and thoroughly reviewed the factorising of polynomials of degree greater than 2.


    I differentiated f(x) =x^n-nx+n-1 in the usual way to get f'(x) = n x^{n-1} - n. Clearly x = 1 is a root of f'(x) and therefore (x - 1) is a factor of f'(x).

    Then I differentiated f(x) = (x-1) h(x) using the product rule. I used the result to show that if (x-1) is a factor of f'(x) then it must be a factor of h(x).


    I have no idea what your mathematics background. It might be that you don't know the product rule. In which case you have to accept as a rule that if f(x) and f'(x) both have a factor of (x - 1), then (x - 1) is a repeated factor of f(x), that is, (x-1)^2 is a factor of f(x).
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  13. #13
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    dont understand . y h(x) + (x-1)h'(x) . how 2 exist?


    f(x) = (x-1) h(x) using the product rule. I used the result to show that if (x-1) is a factor of f'(x) then it must be a factor of h(x).

    at here. i understood that h(x) is quadrant .
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  14. #14
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    Quote Originally Posted by qweiop90 View Post
    [snip]

    dont understand . y h(x) + (x-1)h'(x) . how 2 exist?

    [snip]
    In post #12 I said I used the product rule to get that result. You clearly have not learnt this rule yet. So forget all of that. Please recall:

    Quote Originally Posted by mr fantastic
    It might be that you don't know the product rule. In which case you have to accept as a rule that if f(x) and f'(x) both have a factor of (x - 1), then (x - 1) is a repeated factor of f(x), that is, (x-1)^2 is a factor of f(x).
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