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Rational zero theorem

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There are **multiple options** for p and q, it's not just -6 and 1. p and q are **factors of** 6 and 1.

      a₀=6  ⇒  p ∈ {1,2,3,6}
      aₙ=1  ⇒  q ∈ {1}

    p/q ∈ {1/1, 2/1, 3/1, 6/1}

And all of these can also be negative, so you actually have

    p/q ∈ {+1/1, +2/1, +3/1, +6/1,
           -1/1, -2/1, -3/1, -6/1}

Here's another example: f(x)=2x³-4x²+5x-9

      a₀=9  ⇒  p ∈ {1,3,9}
      aₙ=2  ⇒  q ∈ {1,2}

    p/q ∈ {+1/1, +3/1, +9/1, +1/2, +3/2, +9/2,
           -1/1, -3/1, -9/1, -1/2, -3/2, -9/2}
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According to the fundamental theorem of algebra, the polynomial can be expressed as f(x) = k(x + a)(x + b)…  notice that all the constants, when multiplying first-degree polynomials, are multiplied together to create the constant term of the product, and all the coefficients of x multiplied together to create the leading coefficient. Therefore, in order for a zero to be rational, it must be a ratio of a factor of the constant to a factor of the leading coefficient, since if ax - b = 0, x = b/a. For your example, all the constants in the factored form multiply to 6, and all the leading coefficients multiply to one. Therefore, if its zeroes were all rational, they’d follow the form (x +/- a)(x +/- b)(x +/- c), where a, b, and c are factors of 6.

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