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How can you factorise polynomial by inspection? watch

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    Hi. I remember reading a post in this forum about how to factorise polynomial without long devision but I can't find it anymore. Can someone please direct me to that post or perhaps explain how it can be done?

    Thanks X
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    Try and spot solutions of the polynomial. If x=3 is a solution, then (x-3) is a factor.
    Also, bear in mind that the product of the roots is equal to -q/a, for f(x)= ax^n + bx^(n-1) + ... + px + q.
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    For a worked example, see this post:

    http://www.thestudentroom.co.uk/show...&postcount=456
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    (Original post by calcium878)
    Try and spot solutions of the polynomial. If x=3 is a solution, then (x-3) is a factor.
    Also, bear in mind that the product of the roots is equal to -q/a, for f(x)= ax^n + bx^(n-1) + ... + px + q.
    t's only -q/a for a polynomial of odd order.
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    (Original post by Adje)
    For a worked example, see this post:

    http://www.thestudentroom.co.uk/show...&postcount=456
    That's that one I am looking for. Thank you!
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    (Original post by Totally Tom)
    t's only -q/a for a polynomial of odd order.
    Aye, this is true.

    *rephrases more generally*: the product of the solutions should give the constant of the polynomial (if there isn't one, then one or more of the solutions is zero ), give or take a negative sign.
    So, if it's f(x) = blah blah + 26, then, unless the exam board is being twatty and using fractional solutions, you can count on there being (maybe all of) a 1, 2 and 13 about.
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    (Original post by calcium878)
    Aye, this is true.

    *rephrases more generally*: the product of the solutions should give the constant of the polynomial (if there isn't one, then one or more of the solutions is zero ), give or take a negative sign.
    So, if it's f(x) = blah blah + 26, then, unless the exam board is being twatty and using fractional solutions, you can count on there being (maybe all of) a 1, 2 and 13 about.
    lol and 26...
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    (Original post by calcium878)
    Aye, this is true.

    *rephrases more generally*: the product of the solutions should give the constant of the polynomial (if there isn't one, then one or more of the solutions is zero ), give or take a negative sign.
    So, if it's f(x) = blah blah + 26, then, unless the exam board is being twatty and using fractional solutions, you can count on there being (maybe all of) a 1, 2 and 13 about.
    Or a 26. E.g. x^3 - 26x^2-x+26 has roots 1,-1 and 26.

    As far as fractional solutions, if you have a polynomial

    x^n +a_{n-1}x^{n-1} + ... +a_1 x + a_0 = 0 with all the coefficients being integers, then there are never any fractional roots. Any roots are either integer or irrational.

    (N.B. note the leading coefficient is one, however. The result isn't true if the leading coefficient isn't 1 (or -1)).
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    (Original post by Totally Tom)
    lol and 26...
    Again, true.
    Though it'd be a bit mean for them to expect you to find f(26) = 0 if you didn't realise the whole product of roots thing :p:
 
 
 
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