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Math - polynomials

  1. Nov 21, 2003 #1
    given that f(x) = x^4 - 27x^2 - 14x + 120 can be expressed as

    ( x^2 + a )^2 - ( bx + 7 )^2

    where a,b are constant. find the values of a and b. hence, or otherwise, factorise f(x) completely.




    the value of a and b are -13 and 1 respectively.
    so,

    f(x) = ( x^2 - 13 )^2 - ( x + 7 )^2

    one way to factorise f(x) completely is to substitute value of x so that we can get
    f(x) = 0.

    is there any other easier way since there is a keyword "HENCE" in the question.
    thank you.
     
  2. jcsd
  3. Nov 21, 2003 #2
    f(x) = 0
    <=>
    ( x^2 - 13 )^2 = ( x + 7 )^2
    Taking square roots on both sides:
    |x^2 - 13| = |x + 7|
    This breaks up into 2 separate equations which are just quadratic and, thus, can directly be solved, giving 2 zeroes of f each.
     
  4. Nov 21, 2003 #3

    HallsofIvy

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    I suspect what was intended here was that you use

    a2- b2= (a-b)(a+b).

    Once you know that
    x4 - 27x2 - 14x + 120= ( x2 - 13 )2 - ( x + 7 )2,

    you can continue as
    (x2- 13+ (x- 7))(x2-13-(x-7))
    = (x2+ x- 20)(x2-x- 6)

    Now can you further factor those two factors?
     
  5. Nov 22, 2003 #4

    HallsofIvy

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    "That's Quaint"? :)
     
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