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    This question was in the jan 2009 paper, last question... i swear its a bit hard for c4?...

    You work out that 2t^{3}-5t^{2}-16t-9 = 0
    but then how are you supposed to find t from that?
    in the mark scheme it says 'a realisation that (t+1) is a factor'
    to get (t+1)(2t^{2}-7t-9) = 0 (which you can then work out t from)
    how are you supposed to 'realise' that though?! anyone know?
    cheers

    qp:http://www.freeexampapers.com/FreeEx...IwMDktMDEucGRm
    ms: http://www.freeexampapers.com/FreeEx...ktMDEgTVMucGRm
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    Can't remember the exact name, but the factor theorem(?) states that for a function f(x), if f(a) = 0, then (x-a) is a factor.

    So in this case, since putting t=-1 gives zero, (t--1) i.e. (t+1) is a factor!

    Just do polynomial division and you're there!
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    In C4 if they ask you to solve a cubic in x, plug in x=-2,-1,0,1,2 and you'll probably get a soln. in there.
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    2t^3 - 5t^2 - 16t - 9 = (t + 1)(at^2 + bt + c)

    expand brackets and equate coefficients
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    This is a C1 question?

    A standard method for solving cubics is to try numbers using the factor theorem:
    If f(a) = 0, then (x-a) is a factor.

    Your constant is -9, so you try factors of that:
    1, -1, 3, -3

    Plugging in -1 gives f(t) = 0, so (t+1) is a factor.
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    (Original post by AnonyMatt)
    This is a C1 question?

    A standard method for solving cubics is to try numbers using the factor theorem:
    If f(a) = 0, then (x-a) is a factor.

    Your constant is -9, so you try factors of that:
    1, -1, 3, -3

    Plugging in -1 gives f(t) = 0, so (t+1) is a factor.
    I don't know if it's the same for Edexcel, but with AQA in Core 4 they revisit factor theorom/remainder theorom/division of polynomials. Even though its pretty much the same as in Core 1 :p:
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    (Original post by john1234567)
    2t^3 - 5t^2 - 16t - 9 = (t + 1)(at^2 + bt + c)

    expand brackets and equate coefficients
    There is no need to expand the brackets. If you look at the constant terms above it should be obvious that c=-9 The value of a is 2

    To find  b consider the coefficients necessary to get b so - 16 = b + c and we already know that c is -9
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    (Original post by jlittle)
    Can't remember the exact name, but the factor theorem(?) states that for a function f(x), if f(a) = 0, then (x-a) is a factor.

    So in this case, since putting t=-1 gives zero, (t--1) i.e. (t+1) is a factor!

    Just do polynomial division and you're there!

    ah yeah i remember that, vaguely

    awesome thank you!
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    (Original post by Weeman5872)
    I don't know if it's the same for Edexcel, but with AQA in Core 4 they revisit factor theorom/remainder theorom/division of polynomials. Even though its pretty much the same as in Core 1 :p:
    Oh. :p:

    Well I'm with WJEC, and they revisit differentiation from first principles in FP1, even though it's the same, just with fractions. Suppose this is the same thing really.

    And don't tell anyone but...
    ...I've never divided polynomials. I didn't like it so I stuck to comparing coefficients. :p:
 
 
 
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