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    Hey!

    I've been told to find all roots of f(x) = x^3 -3x^2 - 25x + 75

    How on earth do I do that? Do I factorise first and then use sign diagrams?
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    It's fairly easy to spot a root. Then divide it out and solve the quadratic.
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    Just try and factorize it into three linear factors. You should be able to do this if you can guess, know the factor theorem, and can do algebraic long division.
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    Generally you find 1 by guessing, divide by x - a where a was the root, and then factorise the quadratic you have left. Have you found at least 1 root? (by guessing)
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    Factorize 75. Substitute the factors (incl the negative ones, e.g. -1, -25) of ascending absolute value. When you find, say, f(a) = 0, bingo. It's (x-a) is one of the factor. Then you can use division.
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    (Original post by benwellsday)
    Generally you find 1 by guessing, divide by x - a where a was the root, and then factorise the quadratic you have left. Have you found at least 1 root? (by guessing)
    Erm . . . no. :o:

    I've never done any of this before. It's part of a maths course in economics (it's completely pointless because I am doing joint honours and never actually do any maths after this - I never even apply it) but everyone else has done AS/A level before and I am in open water.
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    (Original post by HappinessHappening)
    Erm . . . no. :o:

    I've never done any of this before. It's part of a maths course in economics (it's completely pointless because I am doing joint honours and never actually do any maths after this - I never even apply it) but everyone else has done AS/A level before and I am in open water.
    Refer to the method I provided above.
    And what model of calculator you used? Perhaps you could input the formula in it.
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    Alright well you may not know about remainder theorem. The only important part is that if f(a) = 0, then x - a divides the polynomial f(x). This is kind of because you get (x-a) when you factorise. To make this more clear, in your example, 5 is a root. f(5) just means
     f(5) = 5^3 - 3(5^2) - 25(5) + 75 = 0 so x - 5 is one of the factors. There are two ways to do the next bit, polynomial long division and equating coeffecients which I think is much easier. Basically you have (x-5) multiplied with some quadratic so that you get  x^3 -3x^2 - 25x + 75 so you write out a general quadratic like this  (x-5)(ax^2+bx+c) = x^3 -3x^2 - 25x + 75 . You can find a and c instantly because ax^2 times x is the only x^3 bit, so a = 1 because you only want 1 x^3. This is similair for c, there are -5c on the left, and on the right there is 75 as a constant, so c = -15. Finding b is slightly more of a pain but it's the same idea. Once you have the quadratic, you then solve that by whatever method you like. Hopefully that makes some sense.
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    (Original post by splee)
    Refer to the method I provided above.
    And what model of calculator you used? Perhaps you could input the formula in it.
    The model is Casio fx-83ES

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    (Original post by benwellsday)
    Alright well you may not know about remainder theorem. The only important part is that if f(a) = 0, then x - a divides the polynomial f(x). This is kind of because you get (x-a) when you factorise. To make this more clear, in your example, 5 is a root. f(5) just means
     f(5) = 5^3 - 3(5^2) - 25(5) + 75 = 0 so x - 5 is one of the factors. There are two ways to do the next bit, polynomial long division and equating coeffecients which I think is much easier. Basically you have (x-5) multiplied with some quadratic so that you get  x^3 -3x^2 - 25x + 75 so you write out a general quadratic like this  (x-5)(ax^2+bx+c) = x^3 -3x^2 - 25x + 75 . You can find a and c instantly because ax^2 times x is the only x^3 bit, so a = 1 because you only want 1 x^3. This is similair for c, there are -5c on the left, and on the right there is 75 as a constant, so c = -15. Finding b is slightly more of a pain but it's the same idea. Once you have the quadratic, you then solve that by whatever method you like. Hopefully that makes some sense.
    That's made alot more sense of it, yes! I think I'll need to practice more using that method but thank you so much!
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    Can calculators solve polynomials now? I can't think of a way that a calculator would be useful for solving a polynomial other than some of the adding up.
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    (Original post by HappinessHappening)
    The model is Casio fx-83ES

    I am not sure if you can input my programmes onto it. It seems it had only 79 bytes of memory but mine takes 139. I think you got to get an fx-50F Plus or fx-3650P.

    http://lpl.hkcampus.net/~lpl-wwk/Cas...ations%201.htm The elementary formula for cubic equations, which can only find real roots. 139 bytes.
    http://lpl.hkcampus.net/~lpl-wwk/Cas...ations%204.htm This one can find complex roots. 160 bytes.
    http://lpl.hkcampus.net/~lpl-wwk/Casio50/Choice.htm Main page for formulas. You may find other equations here.
    http://lpl.hkcampus.net/~lpl-wwk/ Main page of the site

    I'm sorry that the web page is in traditional Chinese - actually 3650P is nearly the standard calculator for HK public exams. But the webpage owner had kindly provided English terms in parenthesis, so I think it won't be too difficult for you to find the formula you want.
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    Hint
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    notice 1,3 and 25, 75

    Solution
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     x^3 - 3x^2 - 25x + 75 = 0

\implies x^2(x-3) - 25(x-3) = 0

\implies (x^2-25)(x-3) = 0

\implies (x-5)(x+5)(x-3) = 0



x=\pm5,x=3


    Edit: Proud of my first latex post. :P
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    (Original post by splee)
    I am not sure if you can input my programmes onto it. It seems it had only 79 bytes of memory but mine takes 139. I think you got to get an fx-50F Plus or fx-3650P.

    http://lpl.hkcampus.net/~lpl-wwk/Cas...ations%201.htm The elementary formula for cubic equations, which can only find real roots. 139 bytes.
    http://lpl.hkcampus.net/~lpl-wwk/Cas...ations%204.htm This one can find complex roots. 160 bytes.
    http://lpl.hkcampus.net/~lpl-wwk/Casio50/Choice.htm Main page for formulas. You may find other equations here.
    http://lpl.hkcampus.net/~lpl-wwk/ Main page of the site

    I'm sorry that the web page is in traditional Chinese - actually 3650P is nearly the standard calculator for HK public exams. But the webpage owner had kindly provided English terms in parenthesis, so I think it won't be too difficult for you to find the formula you want.
    Ah, thank you very much indeed.
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    (Original post by hysterian)
    Hint
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    notice 1,3 and 25, 75

    Solution
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     x^3 - 3x^2 - 25x + 75 = 0

\implies x^2(x-3) - 25(x-3) = 0

\implies (x^2-25)(x-3) = 0

\implies (x-5)(x+5)(x-3) = 0



x=\pm5,x=3


    Edit: Proud of my first latex post. :P
    Wohoo! That's precisely what I got!

    Thanks!
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    (Original post by HappinessHappening)
    Wohoo! That's precisely what I got!

    Thanks!
    Did you get it using a calculator? I have one (Casio don't remember the model) that can solve polynomials, vectors, matrices .............
    PS: I'm not bragging
 
 
 
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