# How can I find the roots of the polynomial $12x^{4}+2x^3+10x^2+2x-2$?

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.

• 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. Jul 14, 2020 at 19:56
• (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 $\Bbb Q$. Jul 14, 2020 at 20:03
• Also, this question must be effectively a duplicate, but a quick search turned up no candidates. Jul 14, 2020 at 20:15

The hint.

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

• ...as the coefficients are all real... Jul 14, 2020 at 14:22

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).

• 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 $x^2 + 1$. Jul 14, 2020 at 19:33

Here, I try to give a way of factorization, which isn't too hard to be noticed:
$$6x^4+x^3+5x^2+x-1$$
$$=5x^4+x^3+5x^2+x+x^4-1$$
$$=x^3(5x+1)+x(5x+1)+(x^2+1)(x^2-1)$$
$$=x(x^2+1)(5x+1)+(x^2+1)(x^2-1)$$
$$=(x^2+1)(6x^2+x-1)$$
$$=(x^2+1)(3x-1)(2x+1)$$