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X2 -little bit of help. watch

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    yeh, i just checked... it doesn't work for another function i made up, but defo works for this. Browny points to who that can work out why.
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    I got it.... it ll only work if its in the format of

    (x - a)^2(x + 2a)

    I think....

    and

    (x + a)^2(x - 2a)
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    (Original post by Frater)
    I got it.... it ll only work if its in the format of

    (x - a)^2(x + 2a)

    I think....

    and

    (x + a)^2(x - 2a)
    You're right it would work if the function was in that form. It would also work if you had:
    f(x).(x+a)^n
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    yeh, well we learnt something new today huh? lol

    Its because it noly touches the root a and does not intersect, just tried it out on "graph" and it works.
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    (Original post by Frater)
    yeh, well we learnt something new today huh? lol

    Its because it noly touches the root a and does not intersect, just tried it out on "graph" and it works.
    To alaborate on that, any polynomial with discriminant equal to zero will follow your method.

    EDIT: I mean a function with a repeated root. Are there discrimants for non-quadratic polynomials?
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    It works for any repeated root, whatever the order. So it would work for, say, (x+1)(x+2)^3, which does actually intersect the x-axis when x = 2.
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    Thats the one.... can't see how this could make maths easier for someone though or help solve an equation, must be some use somewhere.
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    (Original post by DFranklin)
    It works for any repeated root, whatever the order. So it would work for, say, (x+1)(x+2)^3, which does actually intersect the x-axis when x = 2.

    Yup just checked that on graph and it works. Although not for everyone, it seems that there has to be a and 2a .... if that makes sense
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    (Original post by Frater)
    Yup just checked that on graph and it works. Although not for everyone, it seems that there has to be a and 2a .... if that makes sense
    Give an example where it doesn't work.
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    (Original post by notnek)
    Give an example where it doesn't work.
    My bad, jus checked it on graph, it does work.
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    (Original post by Frater)
    Thats the one.... can't see how this could make maths easier for someone though or help solve an equation, must be some use somewhere.
    The converse is often used to check if a polynomial p(x) has repeated zeros: if it does, then p(x) and p'(x) have a common factor, and you can check for that using the Euclidean algorithm without actually needing to factorise p or p'.
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    Cool, never come across that in A level.
 
 
 
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