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It's clear that I can divide by $2$, but I don't know what can I do with $$6x^{4}+x^3+5x^2+x-1$$

Is there any algorithm for it or a trick? I have found the roots by an online calculator but I don't know how can I calculate them. Thank you for your help.

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    $\begingroup$ A general algorithm for factoring quartics $p$ over $\Bbb Q$ is: (1) Check for rational roots; the Rational Root Theorem guarantees that there are only finitely many cases to check (for this polynomial there are only $8$: $\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}$. Sometimes you can reduce the number of cases to check with judicious use of Descartes' Rule of Signs. If you find a root $r$, then $x - r$ is a factor of $p$, and dividing $p$ by $x - r$ using long division reduces the problem to finding a cubic. $\endgroup$ Jul 14, 2020 at 19:56
  • $\begingroup$ (2) If $p$ has no rational roots, then check whether it factors as a product of two quadratics: $A (x^2 + b x + c) (x^2 + d x + e)$, where $A$ is the coefficient of $x^4$ in $p$. Distributing and comparing like terms in $x$ gives a set of 4 (at most) quadratic equations in $b, c, d, e$. If there are no rational solutions, $p$ is irreducible over $\Bbb Q$. $\endgroup$ Jul 14, 2020 at 20:03
  • $\begingroup$ Also, this question must be effectively a duplicate, but a quick search turned up no candidates. $\endgroup$ Jul 14, 2020 at 20:15

3 Answers 3

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The hint.

Easy to see that $i$ is a root, which gives a factor $x^2+1.$

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  • $\begingroup$ ...as the coefficients are all real... $\endgroup$ Jul 14, 2020 at 14:22
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Another hint: In general if you search for rational roots and try inserting $x=p/q$ (irreducible), then $6p^4+p^3 q+ 5 p^2q^2+p q^3 -q^4=0$ implies that $q$ should divide $6$ and $p$ should divide 1. For details look up Rational root theorem in wikipedia.

In the present situation you will find $1/3$, $-1/2$ in this way. If you include the possibility of $p$ being imaginary then you also pick up $\pm i$ (but this is perhaps a bit cheating).

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  • $\begingroup$ Of course, once we know that $\frac{1}{3}, -\frac{1}{2}$ are roots, we can successively divide the given polynomial by the linear polynomials $3 x - 1, 2x + 1$ to compute that the remaining factor is $x^2 + 1$. $\endgroup$ Jul 14, 2020 at 19:33
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Here, I try to give a way of factorization, which isn't too hard to be noticed:
$6x^4+x^3+5x^2+x-1$
$=5x^4+x^3+5x^2+x+x^4-1$
$=x^3(5x+1)+x(5x+1)+(x^2+1)(x^2-1)$
$=x(x^2+1)(5x+1)+(x^2+1)(x^2-1)$
$=(x^2+1)(6x^2+x-1)$
$=(x^2+1)(3x-1)(2x+1)$

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