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    Given that 3+i is a root of the equation fx=0 where

    f(x) = 2x^3+ax^2+bx-10

    (a) find the other two roots of the equation f(x) = 0..

    (b) find the value of a and b..

    (a) If 3+i is a root then I assume 3-i is a root too. How do i find the third root? I tried multiplying (x-3+i) and (x-3-i) but when i factorised that I just got 3+or-i again.. I'm a bit confused.. help?
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    when you get the quadratic you multiply it by a real root (x+c) the will give you f(x)
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    (Original post by Swayum)
    What do you get for (x - 3 + i)(x - 3 - i)?
    x^2 - 6x + 10...
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    Sorry, I don't know if there's something you're MEANT to do for the exam, but I would just look at it now and see that the third factor would obviously be (2x - 1). You need the 2x to get the 2x^3. You need the -1 to get the -10.
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    (Original post by Swayum)
    Sorry, I don't know if there's something you're MEANT to do for the exam, but I would just look at it now and see that the third factor would obviously be (2x - 1). You need the 2x to get the 2x^3. You need the -1 to get the -10.
    Oh ok, so am I right in thinking, to find a and b i subsitute 1/2, 3-i and 3+i and solve?
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    You COULD, but I'd just expand out (x^2 - 6x + 10)(2x - 1) and compare coefficients. Much faster.
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    Product of roots = \frac{10}{2}
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    if product of roots = 5 then multiply out:
    (3+i)(3-) = 9-i^2 = 10
    10 * something = 5
    something = 5/10 = 0.5
 
 
 
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