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Math Help - Finding all the Zeros of a Polynomial Equation (already solved for two zeros)

  1. #1
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    Finding all the Zeros of a Polynomial Equation (already solved for two zeros)

    Hi everyone,

    I cannot seem to figure out how derive the last two zeros of the equation.

    Directions read 'Solve' : 3x^4 - 4x^3 + 23x^2 - 36x - 36 = 0, given that one zero is 2/3

    Using the given zero, I plugged -2/3 into synthetic division which resulted in a remainder of 0, so I changed the equation to:

    (x+2/3)(3x^3 - 6x^2+27x-54) = 0

    Next I randomly chose a factor of 2 and carried out another synthetic division, resulting in a remainder of 0 again. Now the equation looks like:

    (x+2/3)(x-2)(3x^2 + 0x + 27) = 0

    **This is where I am not quite sure how to solve for remaining zeros.

    Do I solve the unsolved portion by pulling out a factor, making it 3(x+3)(x+3)? That would leave me with two zeros of -3...

    If anyone could advise I would greatly appreciate it.

    Thank you.
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  2. #2
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    Re: Finding all the Zeros of a Polynomial Equation (already solved for two zeros)

    Hello, Odonsky!

    There's something wrong with the problem . . . \tfrac{2}{3} is not a zero of the equation!


    \text{Solve: }\:3x^4 - 4x^3 + 23x^2 - 36x - 36 \:=\: 0,\:\text{ given that one zero is }{\color{red}-}\tfrac{2}{3}

    I divided by (3x+2) and got:
    . . f(x) \:=\:(3x+2)(x^3 - 2x^2 + 9x - 18)

    Like you, I saw that x=2 is a zero and divided by (x-2).
    . . f(x) \:=\:(3x+2)(x-2)(x^2 + 9)

    That last factor cannot be factored (in real numbers).
    . . x^2+9\:=\:0 \quad\Rightarrow\quad x^2 \:=\:-9 \quad\Rightarrow\quad x \:=\:\pm\sqrt{-9} \:=\:\pm3i


    The four zeros are: . -\tfrac{2}{3},\;2,\;\pm3i
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    Re: Finding all the Zeros of a Polynomial Equation (already solved for two zeros)

    Thank you Soroban! Your reply is very clear.

    After posting this thread, I worked the latter half of the problem out using the quadratic formula whereas a=3, b=0 & c=27 and arrived at the same solution of +/- 3i.

    Interesting how math works in so many different ways...
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  4. #4
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    Re: Finding all the Zeros of a Polynomial Equation (already solved for two zeros)

    So it doesn't bother you that the whole problem is wrong?
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    Re: Finding all the Zeros of a Polynomial Equation (already solved for two zeros)

    From what I understand the problem has been solved correctly.
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