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Factoring irrational equations

  1. Jan 8, 2005 #1
    Ok, I hope there’s a math wiz out there who can help me.

    I have to factor: [tex]f (x)=3x^4-8x^3-5x^2+16x-5[/tex]

    Just from looking at it, you know that possible values for x are: ±1, ±5/3, ±1/3, ±5

    However, if you plug in these real numbers, none of them work, therefore meaning that x is an irrational number and is complex/imaginary. Note that I’m using synthetic division to solve the equation.

    So, my question would be, how do I solve for irrational numbers? The quadratic equation will probably be required, but I’m still not sure how to get the final answer.

    I’d appreciate an answer along with the steps to solving the problem. This is for extra credit in my class, so we have not been taught yet, and I’d like to get extra credit. I realize I haven't given much time for this equation to be solved, but Id appreciate an answer by Monday.
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  3. Jan 8, 2005 #2


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    Last edited: Jan 8, 2005
  4. Jan 8, 2005 #3


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    You mean the possible rational values...

    Are you sure you're expected to factor it over the complexes? Frequently, when one talks about factoring an integer polynomial, they merely want it factored into other integer polynomials. (So, by this meaning, x^2 - 2 doesn't factor, although it does over the reals)

    So, you know it doesn't have any linear factors (and thus doesn't have any cubic factors) -- you need to check if it has quadratic factors.

    If you really are intent on solving it, if factoring into quadratic factors doesn't work, you'll either have to resort to the quartic formula (very ugly, though it might be less ugly if you work through the derivation, sort of like completing the square instead of using the quadratic formula directly) or numerical approximation.
  5. Jan 8, 2005 #4


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    Zurtex: there's a theorem, I think called the rational root theorem, that if an integer polynomial has a rational root, the numerator divides the constant term, and the denominator divides the leading term.

    For a quick heuristic way to see it, suppose (ax - b) was a factor of the equation... how do a and b relate to the coefficients of the polynomial?
  6. Jan 8, 2005 #5


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    I'm not a math wiz,yet i'll give it a try.
    Mutiply,equate sides and get the system of equations
    [tex] a+3c=-8 [/tex](2)
    [tex] 3d+b+ac=-5 [/tex](3)
    [tex] ad+bc=16 [/tex] (4)
    [tex] bd=-5 [/tex] (5)

    ,which is very solvable.You can fix another parameter and obtain a system 2 with 3.But that would be lucky,as u'd get 3 eq.with 2 unknowns,and the sytem of eq.must be compatible.E.g.fixing "a=1" induces "c=-3" and the system of 3 eqns.with only 2 unknowns ("b" and "d") which would follow from (3),(4) and (5) would be incompatible.
    Since the polynomial is factored as a product of 4 monoms each of them containing a root as a free term,the problem is actually of chosing the method:you either use the algorithm to solve this quartic,either solve the system above.

  7. Jan 8, 2005 #6
    well, for problems like these, this is what I do (I'll try my best at explaining):

    Take the last number (the -5) and take all the factors that make this up (±1, ±5). Then do the same for the leading coefficient (3), which gives you (±1, ±3). Put the factors of the -5 over the factors of 3 to get this:

    [tex]\frac{±1, ±5}{±1, ±3}[/tex]-Note that the numbers each have the "±" sign, but I'm not sure how to put it in the LaTeX system.

    From this, you can see these:
    [tex]\frac{±1}{±1}[/tex], [tex]\frac{±1}{±3}[/tex], [tex]\frac{±5}{±3}[/tex], [tex]\frac{±5}{±1}[/tex] (also note that these are both plus and minus). This gives you possible real number solutions (±1, ±5/3, ±1/3, ±5)

    Take these numbers, and put them into a synthetic division problem. I really have no idea how Latex would handle this, so I'll write it myself:

    [ ] 3 -8 -5 +16 -5


    In the bracket's, you choose one of the real numbers you've found from above. Then, drop 3 down below the line and multiply it by whatever number is in the bracket (1 for example). Take that multiplied number (which would be 3) and put it under the -8, then add the -8 and 3 to get -5, which you put below the line. Then, multiply the -5 by the number in the brackets and keep doing this over and over.

    When you come to the final answer below the -5, it should be a "0" which tells you that the number in the bracket is a possible value of X. Then from here, you'd continue to write out the equation to find the answer. However, if "0" does not appear, then the number in the brackets is not a value of x, so you must choose from the list of numbers (the ±1, ±5/3, ±1/3, ±5). However, after going through every possible number in that list, none ever got to "0" which means that the equation does not have any rational numbers (so the answer must be irrational). This is the part I didn't know how to solve for.

    I tried factoring, which was tough in my opinion, but you get: (3x^2+x-5)(x^2-3x+1)

    Now from this, you can use the quadratic equation on both sets of numbers. To finish the factorization, use the Quadratic Formula to find the zeroes of the two quadratics.

    So, to sum this all up, the fourth-power polynomial is the product of two quadratic equations.

    I appreciate the help :smile:
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