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Determine an equation of p(x)

  1. Sep 24, 2011 #1
    1. The problem statement, all variables and given/known data

    Determine the equation in simplied form for the family of quartic functions with zeroes of..

    5 (order 2) and -1± 2√ 2



    2. Relevant equations



    3. The attempt at a solution

    so.. (x-5)^2 (x-1+2√2) (x-1-2√2)

    (x-5) (x-5) (x-1+2√2) (x-1-2√2)

    Would be all of it expanded, but I dont know how to foil on this.. theres 2 terms on the left side brackets and 3 terms on the right side brackets.. (x) (-1) (2√2) , I know that X will become x^4, but i dont get how to continue using foil, can anyone explain?

    We never learned this.. so.. i also know that the plusminus root is (x+1)^2 + 8 if simplified into that form..
     
  2. jcsd
  3. Sep 24, 2011 #2

    eumyang

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    This is wrong. It should be
    (x - 5)^2 [x - (-1+2√2)][x - (-1-2√2)]
    = (x - 5)(x - 5)[x + 1 - 2√2][x + 1 + 2√2]

    FOIL the first two binomials, and then FOIL the last two binomials. In FOILing the last two binomials, it may be helpful to rewrite like this:
    [x + 1 - 2√2][x + 1 + 2√2]
    = [(x + 1) - 2√2][(x + 1) + 2√2]
    .. and then use the sum+difference pattern (a - b)(a + b) = a2 - b2.

    Then, multiply the two resulting trinomials (see http://www.purplemath.com/modules/polymult3.htm" [Broken] if you don't know how).

    This is also wrong (probably because you were missing signs earlier. It should be
    (x+1)^2 - 8.
     
    Last edited by a moderator: May 5, 2017
  4. Sep 24, 2011 #3
    wat? explain... why are you distributing random negetives into those brackets
     
  5. Sep 24, 2011 #4

    eumyang

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    Because if a is a root of a polynomial equation f(x) = 0, then
    (x - a)
    is a factor of that polynomial. So there is a negative outside the roots -1+ 2√ 2 and -1 - 2√ 2:
    (x - 5)(x - 5)[x - (-1 + 2√ 2)][x - (-1 - 2√ 2)]
     
  6. Sep 24, 2011 #5
    okay..? so you factor inwards...? like (x+1 -2√ 2) and (x+1 +2√ 2) ?
     
  7. Sep 24, 2011 #6
    any1?
     
  8. Sep 24, 2011 #7

    SammyS

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    What do you mean by "factor inwards" ?

    Also, eumyang basically gave you the next step.
     
  9. Sep 24, 2011 #8

    HallsofIvy

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    The simplest thing to do with complex conjugate terms is to use [itex](a- b)(a+ b)= a^2- b^2[/itex].

    If [itex]-1+ 2\sqrt{2}[/itex] and [itex]-1- 2\sqrt{2}[/itex] then [itex](x- (-1+2\sqrt{2})[/itex] and [itex](x- (-1-2\sqrt{2}))[/itex] are factors.

    [tex]((x- 1)- 2\sqrt{2})((x-1)+ 2\sqrt{2})= (x- 1)^2- (2\sqrt{2})^2= x^2- 2x+ 1- 8= x^2- 2x- 7[/tex].

    It shouldn't be too difficult to multiply [itex](x^2- 10x+ 5)(x^2- 2x- 7)[/itex]
     
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