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How do you the simplified polynomial with integer coefficients and the following root(s):

$\sqrt3+4i$

I think there is something related to conjugates, but I do not know how to do this.

Due to the complex conjugates theorem,

$$((x-\sqrt3)-4i)((x-\sqrt3)+4i) = (x-\sqrt3)^2+16=x^2-2\sqrt3x+19$$ Here, there is a problem since there are no integer coefficients. I tried squaring the expression $x^2-2\sqrt3+19$ as I thought maybe $\sqrt3+4i$ and $\sqrt3-4i$ were double roots, but this was useless, since it produced more irrational solutions.

Does this mean that the expression with $\sqrt3+4i$ and $\sqrt3-4i$ as its roots does not have rational solutions? If not, how would I go about approaching this problem?

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You are on the right track. The complex conjugate of $\sqrt3+4i$ is $\sqrt3-4i$ so we set the polynomial $$((x-\sqrt3)+4i))((x-\sqrt3)-4i))=(x-\sqrt3)^2-16i^2=x^2-2\sqrt3x+19=0$$ Now rearrange the polynomial $$2\sqrt3x=x^2+19$$ and on squaring, $$12x^2=x^4+38x^2+361$$ so the answer is $$\boxed{x^4+26x^2+361=0}$$

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  • $\begingroup$ That’s not the answer. It should be $$x^4 + 26x^2 + 361 = 0.$$ Not $26x$. Now let $u = x^2$ and the trinomial becomes $$u^2 + 26u + 361 = 0.$$ Using the quadratic formula, we get $$u = \frac{-26\pm \sqrt{676 - 1444}}{2} = -13 \pm \frac{\sqrt{-768}}{2}.$$ $\endgroup$ – Mr Pie Feb 4 '18 at 3:24
  • $\begingroup$ $$\begin{align} &= -13\pm i\sqrt{192} \\ x &= \pm\sqrt{-13 \pm i\sqrt{192}}\end{align}$$ such that the $\pm$ operations are independent of each other, giving you four solutions of $x$ in total, as desired. $\endgroup$ – Mr Pie Feb 4 '18 at 3:30
  • $\begingroup$ Thanks. Corrected typo. $\endgroup$ – TheSimpliFire Feb 4 '18 at 10:04
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You can use also the following identity. $$(a+b+c)(a+b-c)(a+c-b)(b+c-a)=2(a^2b^2+a^2c^2+b^2c^2)-a^4-b^4-c^4,$$ where $a=x$, $b=-\sqrt3$ and $c=-4i$.

Now, since $a+b+c=0$, we obtain: $$2(3x^2-16x^2-48)-x^4-9-256=0$$ or $$x^4+26x^2+361=0.$$

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You can try to eliminate the surd and $i$ by isolating terms and squaring the equation $x=\sqrt{3}+4i$.

Method 1: eliminating the surd then the imaginary

$\begin{align}(x-4i)^2=3 &\implies x^2-8ix-19=0 \\&\implies (x^2-19)^2=-64x^2 \\&\implies x^4+26x^2+361=0\end{align}$

Method 2: eliminating the imaginary then the surd

$\begin{align}(x-\sqrt{3})^2=-16 &\implies x^2-2x\sqrt{3}+19=0 \\&\implies (x^2+19)^2=12x^2 \\&\implies x^4+26x^2+361=0\end{align}$

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