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Solve for the rational roots of

$$\begin{align*}f(x)=x^4-5x^3+11x^2-16x+12\tag{1}\end{align*}$$

I know the rational roots are factors of 12, so just try $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12$ one by one?

Bad thing is try many times, I've tried $1$, and $-1$ and $2$ is.(Is any rules in trying? $1$ first or $-1$ first, or $2$ first?)

Lucky thing is $2$ is one root, and multiplicity of $x-2$ is $2$.

So by method of undetermined coefficients, we can get the factor.

$$\begin{align*}x^2-x+3\tag{2}\end{align*}$$

Shoud I show $(2)$ is irreducible over $Q$, and how? calculate values when $x$ takes over $\pm 3,\pm 4,\pm 6,\pm 12$?

forlorn
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  • It is irreducible, no rational roots. Use the Quadratic Formula. Or try the only candidates for roots $\pm 1$, $\pm 3$. – André Nicolas Aug 09 '13 at 06:34
  • @AndréNicolas Why there is no need to check 4,6,12? If I use Quadratic Formula, then it is something like this – forlorn Aug 09 '13 at 11:14
  • You have arrived at $x^2-x+3$, and want to find out whether it is irreducible. Like any quadratic or cubic, it is irreducible iff it has no rational root. And any rational root of $x^2-x+3$ must divide the constant term of this polynomial, so it must divide $3$. – André Nicolas Aug 09 '13 at 14:08

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The next step is indeed to show that $x^2 - x + 3$ is irreducible over the rationals; but you don't need to look at that whole list of possible roots. You just need to look at what the rational root theorem tells you about this new, smaller polynomial: so you only need to check $1, -1, 3, -3$.

Don Laackman
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