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How can I show that $\mathbb R [x]/(x^5+x-3)$ is not an Integral domain?

To prove that it is not an Integral domain at first I have show that $(x^5+x-3)$ this ideal is not a prime ideal as $\mathbb R [x]$ is a commutative ring with unity. For this I have to show $a(x)b(x)$ product of two polynomials belongs to the ideal $(x^5+x-3)$ but $a(x)\ \&\ b(x) \notin (x^5+x-3)$.

I can't find such $a(x)\ \&\ b(x)$?

Is there any other method to do this?

Please suggest some edit.

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    $\begingroup$ You don't really need to show factors, just show that they exist. Let $r$ be any root of $x^5+x-3$, then either $x-r$ is a (real polynomial) factor or $(x-r)(x-\overline{r})$ is one. $\endgroup$ – conditionalMethod Nov 8 '19 at 10:24
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    $\begingroup$ So, as long as the degree is $\geq 3$ it will factor over $\mathbb{R}$. $\endgroup$ – conditionalMethod Nov 8 '19 at 10:31
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Since $x^5+x-3$ is an odd dgree polynomial, it has some real root $r$. So, you can write it as $(x-r)\times q(x)$ for some polynomial $q(x)$ with degree $4$ and therefore in $\mathbb R[x]/(x^5+x-3)$ you have $(x-r)\times q(x)=0$.

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Any real polynomial of odd degree has a real root. In particular, $f=x^5+x-3$ is not irreducible and hence not prime. Note that $\Bbb R[X]$ is a factorial ring. But $\Bbb R[X]/(f)$ is an integral domain if and only if $(f)$ is a prime ideal.

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