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What is the simplest way to get the real solutions for this equation?

$$ x^{4}-2x+1=0 $$

I can do it and also ask for a step-by-step solution to Wolfram Alpha, but I was wondering if someone has a simpler way...

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Rational root test tells us that if there is a rational root $p/q$, then $p$ divides free coefficient $a_0 = 1$ and $q$ divides leading coefficient $a_4 = 1$. Thus, the only possible rational roots are $\pm 1$. Quick check confirms that $1$ is a root, but $-1$ is not. Thus, $x^4-2x+1$ is divisible by $x-1$ and the corresponding factorization is $$ x^4-2x+1 = (x-1)(x^3+x^2+x-1) $$

Unfortunately, $x^3+x^2+x-1$ doesn't have any rational roots (so, it is in fact irreducible over $\mathbb Q$) and to find the roots, you can follow Wiki's article on cubic function.

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Notice that the sum of coefficients is 0, so $1$ is a solution. Divide by $x-1$ and simplify.

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  • $\begingroup$ And if you can't immediately see that the sum of coefficients is zero, you can always use the rational root theorem to note that the only possible rational values to test are $\pm1$. The irrational roots are going to be difficult-to-impossible to get any grasp on, so... $\endgroup$ – Steven Stadnicki Aug 31 '18 at 1:14
  • $\begingroup$ Thank you for helping! If you're interested check my follow up question: math.stackexchange.com/questions/2900232/… $\endgroup$ – Igaturtle Aug 31 '18 at 2:10
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Note that sum of coefficients are $0$ so $x=1$ is a real root. Then divide by $x-1$ and focus on the quotient polynomial for another real root.

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