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Math Help - Resolve into factors

  1. #1
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    Resolve into factors

    Resolve into factors :
    1) x^2 +2(a^2 + b^2) + 3ax - b(3x + 5a)
    2) x^4 + 6x^3 + 4x^2 - 15x + 6
    Thanking you
    Regards
    Sreedhar
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  2. #2
    Senior Member I-Think's Avatar
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    Rearranging
     1)  x^2 +2(a^2 + b^2) + 3ax - b(3x + 5a)

    x^2+2a^2+2b^2+3ax-3bx-5ab

    x^2+3(a-b)x+(2a^2-5ab+2b^2)

    Now using the quadratic formula
    \frac{3(b-a)\pm{\sqrt{9a^2-18ab+9b^2-8a^2+20ab-8b^2}}}{2}

    Should be able to solve from there, solve for x and these are your factors

    2) x^4 + 6x^3 + 4x^2 - 15x + 6
    This equation is in its simplest form.
    Any polynomial whose coefficients are integers will, if it has an integer solution, have an integer solution that is a factor of the coefficient of the zeroth order term i.e. the constant. For example look at the general quartic:

    a x^4 + b x^3 + c x^2 + d x + e = 0

    where the coefficients a through e are integers. Well if there is an integer solution to this then it will be a factor of e.
    No factor of 6 makes this equation, hence it's in its simplest form
    Last edited by I-Think; December 7th 2009 at 06:48 AM.
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  3. #3
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    Quote Originally Posted by yshridhar View Post
    [snip]
    2) x^4 + 6x^3 + 4x^2 - 15x + 6
    Thanking you
    Regards
    Sreedhar
    Let u = x^2 + 3x. Then your expressin becomes u^2 - 5u + 6 = (u - 3)(u - 2) = (x^2 + 3x - 3)(x^2 + 3x - 2). The factors are irreducible quadratic factors.
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