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Math Help - third-degree polynomial integer coefficients

  1. #1
    hsmath1918
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    third-degree polynomial integer coefficients

    I have this equation x^3 + ax^2 + bx + c
    I know x = 3/(7^(1/3) - 2) is a solution.
    How can I find a, b, c when they are all integers?
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  2. #2
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    Quote Originally Posted by hsmath1918 View Post
    I have this equation x^3 + ax^2 + bx + c
    I know x = 3/(7^(1/3) - 2) is a solution.
    How can I find a, b, c when they are all integers?
    You have that x = \frac{3}{7^{1/3} - 2} is a solution.

    So,
    \frac{3}{x} = 7^{1/3} - 2

    \frac{3}{x} + 2 = 7^{1/3}

    Cube both sides,

    \frac{27}{x^3} + \frac{54}{x^2} + \frac{36}{x}+ 8  = 7

    Multiply by x^3 thus:

    27 + 54x+36x^2+8x^3 = 7x^3

    Thus, x^3 + 36x^2 + 54x+27=0
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  3. #3
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    Hello, hsmath1918!

    I have this equation: . x^3 + ax^2 + bx + c \:=\:0

    I know x = \frac{3}{\sqrt[3]{7} - 2} is a solution.

    How can I find a,\,b,\,c when they are all integers?

    There may be many cubic equations with that solution . . . or not.
    Here is one of them . . .

    We have: . x \:=\:\frac{3}{\sqrt[3]{7} -2}\quad\Rightarrow\quad\left(\sqrt[3]{7} - 2\right)x\:=\:3

    . . \sqrt[3]{7}x - 2x\:=\:3\quad\Rightarrow\quad \sqrt[3]{7}x \:=\:2x + 3


    Cube both sides: . \left(\sqrt[3]{7}x\right)^3\:=\:(2x + 3)^3\quad\Rightarrow\quad7x^3\:=\:8x^3 + 36x^2 + 54x + 27

    Therefore, an equation is: . x^3 + 36x^2 + 54x + 27\;=\;0

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