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Finding zeros of a quartic function

  1. Jun 12, 2010 #1

    DTA

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    1. The problem statement, all variables and given/known data

    The problem states:

    Consider the function f(x)=x4-8x3+2x2+80x-75

    a) Verify that x-1 and x-5 are factors
    b) Find the remaining factors of f(x)
    c) List all real zeros

    2. Relevant equations

    I did synthetic division to prove that 1 and 5 are factors, yet I'm having trouble figuring out how to get the remaining zeros.


    3. The attempt at a solution

    I tried splitting up the equation and factoring out from both sides and got:

    x3(x-8)+2x(x-40)-75

    And I just got lost from here. I'm blanking on how to find the zeros besides just choosing a bunch of numbers and doing synthetic division a bunch of times.

    Thanks for any help!
     
  2. jcsd
  3. Jun 12, 2010 #2
    If x-1 and x-5 are factors, what do f(1) and f(5) equal, and why?
     
  4. Jun 12, 2010 #3

    vela

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    You might also find it helpful to use the results of the synthetic division to get a polynomial that's easier to factor.
     
  5. Jun 12, 2010 #4

    DTA

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    They both equal zero. I already understood that part, but my teacher is asking for any remaining factors of f(x). I used my graphing calculator and saw f(-3) is another zero, but I don't know how to factor f(x) to algebraically find f(-3) as a zero.
     
  6. Jun 12, 2010 #5
    Since you know that x-1 and x-5 are factors of the function f(x)=x^4-8x^3+2x^2+80x-75 then there exist a function, call it p(x) which as the property that if multiplied with the two factors it yields the function f(x). In other words:

    (Assuming x-1 and x-5 are factors)

    f(x) = x4 - 8x3 + 2x2 + 80x - 75
    ->
    f(x) = p(x) * (x-1) * (x-5)

    You will be able to find p(x) by dividing f(x) with (x-1)(x-5).

    Now you utilize that p(x)'s factors are also factors of the 'original' function, f(x). And you get the factors for p(x), obviously, from solving the equation p(x) = 0.

    (I'm very tired and English is not my native language, so I hope I was able to provide you with some hints)
     
  7. Jun 12, 2010 #6

    vela

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    Synthetic division gives you both a remainder and a quotient. Say you verified the x-5 case first. The bottom row of numbers you got would be 1, -3, -13, 15, 0. The final zero tells you that x-5 divided evenly into f(x); the first four numbers give you the quotient. So you have [itex]f(x)=(x-5)(x^3-3x^2-13x+15)[/itex]. Now if you do synthetic division of [itex]x^3-3x^2-13x+15[/itex] by x-1, you'll be left with a quadratic, which you can hopefully factor by trial and error or by inspection.
     
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