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    (Original post by Clarity Incognito)
    It seems like you're being led on a wild goose chase with some of these methods. Seriously, just look up factor theorem.
    I know how to use the factor theorem. But how would I show that I am factorising? I would still need these general equations right?

    Thanks
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    Are we all suggesting that cannot be divided by (x-3)????????

    seriously?
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    (Original post by 2710)
    Look, stop being a smartass. I do not need to find a root first, as most of these posters have said. I can use a general equation and then compare coefficients. I did not have to find the 'root' for that, I get the root at the end. If you are gonna be sarcastic like that, id prefer if you did not post
    Look, the only way you can possibly factorise a cubic is by finding a root of the equation first. So, for example, finding a number that divides into -27 (somebody listed them above) to substitute for x.

    I was trying to help but never mind.
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    (Original post by TheNack)
    You don't really have to do that with the formulae I showed you. In the formulae I suggested, you have a linear expression to solve (a+-b) and the quadratic expression to solve. That should be pretty straightforward, at least moreso than trying to solve a cubic directly.
    Yes I know, I was just caught up in the heat and giving him an example of how i don't have to find the root first.
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    (Original post by 2710)
    I know how to use the factor theorem. But how would I show that I am factorising? I would still need these general equations right?

    Thanks
    By long division. You divide x^3 -27 by (x-3) and VOILA, you have a quadratic and you have shown how you got it.
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    If you have something in the form x^n-a^n it's fairly obvious that x-a will be a factor since in both cases setting x=a it evaluates to zero. You can work the rest out by polynomial long division or by setting x^3-a^3 = (x-a)(x^2+px+q) and comparing coefficients, or you could just memorise the general result:
    x^3-a^3 = (x-a)(x^2+ax+a^2)

    Even more generally,
    x^n-a^n = (x-a)(x^{n-1} + x^{n-2}a + \cdots + x^{n-k}a^{k-1} + \cdots + xa^{n-2} + a^{n-1})

    Or more concisely,
    \displaystyle x^n-a^n = (x-a)\sum_{r=1}^{n-1} x^{n-r}a^{r-1}
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    Ok , some people on here need a serious Maths revision.

    x^3 - 3^3 = (x-3)(x^2 + 3x + 3^2)
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    (Original post by nuodai)
    If you have something in the form x^n-a^n it's fairly obvious that x-a will be a factor since in both cases setting x=a it evaluates to zero. You can work the rest out by polynomial long division or by setting x^3-a^3 = (x-a)(x^2+px+q) and comparing coefficients, or you could just memorise the general result:
    x^3-a^3 = (x-a)(x^2+ax+a^2)

    Even more generally,
    x^n-a^n = (x-a)(x^{n-1} + x^{n-2}a + \cdots + x^{n-k}a^{k-1} + \cdots + xa^{n-2} + a^{n-1})

    Or more concisely,
    \displaystyle x^n-a^n = (x-a)\sum_{r=1}^{n-1} x^{n-r}a^{r-1}

    lol
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    one factor is obviously x=3.
    So gives:

    (x-3)(x^2+3x+9)
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    (Original post by 2710)
    Look, stop being a smartass. I do not need to find a root first, as most of these posters have said. I can use a general equation and then compare coefficients. I did not have to find the 'root' for that, I get the root at the end. If you are gonna be sarcastic like that, id prefer if you did not post
    Wow jeez, you're being kind of a douche. The most sure fire way of factorising a cubic is to find a root and then do algebraic division...
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    Or (xrootx + 3root3)(xrootx - 3root3)
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    (Original post by JoshFlack)
    Wow jeez, you're being kind of a douche. The most sure fire way of factorising a cubic is to find a root and then do algebraic division...
    Look, I already explained that and he was not cooperating, did you even read the posts? I wanted help, not patronisation.

    And thanks guys, I got it now
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    Hi, 2710. I feel I ought to step in and defend Skadoosh... he was not being patronising or sarcastic. It is quite true that, if you are at A-level standard (which I presume you are), then in order to factorise a cubic (which means find all three factors) you do indeed have to know one first. Be careful in saying you "didn't have to find the root first". There IS no "the root", but there ARE "three roots". And you can find the other two if you know one already, but you do have to know one. If you insist you don't, show me your complete working and I'll show you where you put in that you knew one already.

    In general if x=a is a root of the cubic f(x), then f(a)=0, and (x-a) is a factor. You can then do polynomial division, to show that the cubic is a product of (x-a) with a quadratic, and you can find the quadratic. Then you factorise the quadratic to get the other two roots.

    For x^3-27=0, obviously x=3 is a root, therefore (x-3) is a factor. Dividing x^3-27 by x-3 gives you x^2+3x+9. That quadratic has two roots but they're complex. As for how you would "show that you're factorising", you can either lay it out in the way that polynomial long division is usually laid out, or you could just write " x^3-27 = (x-3)(x^2+3x+9) ". The latter is 'polynomial short division', and personally I find it easier than the 'long' variety.

    Incidentally, Skadoosh later said "the only way you can possibly factorise a cubic is...", which is true enough at A-level but not true in general. Just like there's a quadratic formula, so there's a cubic formula. Some Italian blokes found it in the 16th century. But you don't need to know about it at A-level. Scoot off to wikipedia and you'll see why!
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    (Original post by Nancarrow)
    Incidentally, Skadoosh later said "the only way you can possibly factorise a cubic is...", which is true enough at A-level but not true in general. Just like there's a quadratic formula, so there's a cubic formula. Some Italian blokes found it in the 16th century. But you don't need to know about it at A-level. Scoot off to wikipedia and you'll see why!
    The method of inspection is so much easier than using the cubic formula.

    Like seriously, it requires an unnecessary amount of algebraic manipulation! I guess it's there for completion but I don't think I've ever been in a situation where is has been quicker...probably because the polynomials given at a level are nice integer typesl.
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    >__> Jeez, I wasn't so much debating about finding the root first, but more his tone of voice. I might have overreacted, meh.

    Nancarrow, I kinda skimmed your post, you said that you'd help me point out where I use the first root to get the others?

    Ok here:

      a^3 - b^3 = (a-b)(a^2 +ab +b^2)

    x^3 - 3^3 = etc

    Also for other cubics, you can take out a factor of x, and then factorise the remaining quadratic to get all three roots. That way, you didn't have to have the first root also.

    No hate.

    Thanks
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    Follow Nuodai's advice.

    x^2-y^2 = (x-y) (x+y)
    x^3-y^3 = (x-y) (x^2+x y+y^2)
    x^4-y^4 = (x-y) (x^3+x^2 y+x y^2+y^3)

    \vdots

    x^n-y^n = (x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots+xy^{n-1}+y^{n-1})

    \displaystyle{x^n-y^n=(x-y)\left(\sum^{n-1}_{k=0}{x^ky^{n-1-k}}\right)} \forall{n}\in \mathbb{N}.
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    (Original post by 2710)
    Ok here:

      a^3 - b^3 = (a-b)(a^2 +ab +b^2)

    x^3 - 3^3 = etc
    By taking a factor of (x-3) out of (x^3 - 27) you are recognising that x=3 is a root.

    Also for other cubics, you can take out a factor of x, and then factorise the remaining quadratic to get all three roots. That way, you didn't have to have the first root also.
    Yes you did. When you took a factor of x out, you were saying (x-0) is a factor, in other words, x=0 is a root.

    (And of course, you cannot do that for ALL cubics, only ones where the constant term is zero)

    ANY time you solve a cubic by writing it as a product of a linear factor (x-a) and a quadratic, you have found the root x=a whether you realise it or not.

    And Clarity Incognito, it is of course much easier to find one root by inspection, provided that first root is easy to find! In A-level this is always the case. They do not expect you to use the general cubic formula, nor should they, it's really only of historical interest, to show how S-M-R-T guys like Tartaglia and Cardano really were. They solved the quartic as well! Much later on some guys showed that no such formula can even exist for quintic or higher-degree polynomials, which is a much more interesting result.
 
 
 
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