Can you find all quartic polynomials with real, rational coefficients having $2 - i \sqrt{3}$ $2-isqrt(3)$ and $\sqrt{2} + 1$ $sqrt2 +1$ as two of the zeros?

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Can you find all quartic polynomials with real, rational coefficients having $2 - i \sqrt{3}$ $2-isqrt(3)$ and $\sqrt{2} + 1$ $sqrt2 +1$ as two of the zeros?

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Answer 1 · 2017-04-02T08:26:58

$f (x) = k (x^{4} - 6 x^{3} + 14 x^{2} - 10 x - 7)$ $f(x) = k(x^4-6x^3+14x^2-10x-7)$

for any rational $k \neq 0$ $k != 0$

Explanation:

Note that:

$(x - α) (x - β) = x^{2} - (α + β) x + α β$ $(x-alpha)(x-beta) = x^2-(alpha+beta)x+alphabeta$

So given $α$ $alpha$ of the form $a + b \sqrt{c}$ $a+bsqrt(c)$ with $a, b, c$ $a, b, c$ non-zero rational, we need $β$ $beta$ to be a $a - b \sqrt{c}$ $a-bsqrt(c)$ in order to make both $α + β$ $alpha+beta$ and $α β$ $alphabeta$ rational.

In our example, that means that in addition to the given zeros:

$2 - i \sqrt{3}$ $2-isqrt(3)" "$ and $\sqrt{2} + 1$ $" "sqrt(2)+1$

we must have the conjugate zeros:

$2 + i \sqrt{3}$ $2+isqrt(3)" "$ and $- \sqrt{2} + 1$ $" "-sqrt(2)+1$

So the simplest polynomial with rational coefficients and these zeros is:

$(x - 2 - i \sqrt{3}) (x - 2 + i \sqrt{3}) (x - 1 - \sqrt{2}) (x - 1 + \sqrt{2})$ $(x-2-isqrt(3))(x-2+isqrt(3))(x-1-sqrt(2))(x-1+sqrt(2))$

$= ({(x - 2)}^{2} - {(i \sqrt{3})}^{2}) ({(x - 1)}^{2} - {(\sqrt{2})}^{2})$ $= ((x-2)^2-(isqrt(3))^2)((x-1)^2-(sqrt(2))^2)$

$= ((x^{2} - 4 x + 4) + 3) ((x^{2} - 2 x + 1) - 2)$ $= ((x^2-4x+4)+3)((x^2-2x+1)-2)$

$= (x^{2} - 4 x + 7) (x^{2} - 2 x - 1)$ $= (x^2-4x+7)(x^2-2x-1)$

$= x^{4} - 6 x^{3} + 14 x^{2} - 10 x - 7$ $= x^4-6x^3+14x^2-10x-7$

Any quartic with rational coefficients and these zeros will be a rational multiple of this one.

So the possible quartics are:

$f (x) = k (x^{4} - 6 x^{3} + 14 x^{2} - 10 x - 7)$ $f(x) = k(x^4-6x^3+14x^2-10x-7)$

for any rational $k \neq 0$ $k != 0$

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Can you find all quartic polynomials with real, rational coefficients having $2 - i \sqrt{3}$ $2-isqrt(3)$ and $\sqrt{2} + 1$ $sqrt2 +1$ as two of the zeros?

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Explanation:

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