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If the two solutions of a polynomial are $4$ and $6 - \sqrt{7}$, how would you solve for the polynomial?

To start would you write the expanded form as $(x-4)$ and $(x-(6-\sqrt{7})$ and then multiply?

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  • $\begingroup$ Yes, that gives you a polynomial with those roots. $\endgroup$ – Alex J Best Nov 14 '18 at 23:47
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Yes. The polynomials for with exactly those two roots are precisely those of the form $c(x-4)^a(x-(6 - \sqrt{7}))^b$, for $a, b \geq 1$ integers and $c$ any real number. With further restrictions (integer polynomials, the degrees of the roots, whatever else), you could narrow that down, potentially to a single solution.

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  • $\begingroup$ For what it's worth: You can't have $6-\sqrt7$ be a root of a polynomial with integer (or rational) coefficients without also having $6+\sqrt7$ as a root. $\endgroup$ – Arthur Nov 14 '18 at 23:50
  • $\begingroup$ Exactly what I was thinking, that's just what was provided. $\endgroup$ – Critical Gaming Nov 15 '18 at 1:08

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