# Find integers c and d such that the equation x^3+cx+d=0 has 1+sqrt(3) as one of its roots.

Find integers c and d such that the equation ${x}^{3}+cx+d=0$ has $1+\sqrt{3}$ as one of its roots.
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Anaya Gregory
Given an equation ${x}^{3}+b{x}^{2}+cx+d$ that factors to $\left(x-{r}_{1}\right)\left(x-{r}_{2}\right)\left(x-{r}_{3}\right)$,
$b={r}_{1}+{r}_{2}+{r}_{3}$
$c={r}_{1}{r}_{2}+{r}_{2}{r}_{3}+{r}_{3}{r}_{1}$
$d={r}_{1}{r}_{2}{r}_{3}$
For our problem, we'll say that ${r}_{1}=1+\sqrt{3}$
In order for d to be an integer, either ${r}_{2}$ or ${r}_{3}$ must be $\left(1-\sqrt{3}\right)$ because we have to eliminate theradical in the same way we would if we were rationalizing thedenominator of a fraction.
So far we have ${r}_{1}=1+\sqrt{3},{r}_{2}=1-\sqrt{3}$, and we know that ${r}_{1}+{r}_{2}+{r}_{3}=0$. Using these facts together you should be able to figure out what ${r}_{3}$ is and then plug into the above formul as to find c and d.