# Why doesn't the Rational Root Theorem work on this cubic?

I tried using the Rational Root Theorem on $x^3-6x^2+4x-5=0$. However, I could not find a rational root of the cubic. When I plugged the cubic into Wolfram alpha, it yielded a very messy real solution:

How do you tell whether the RRT doesn't work on a particular cubic of the form $ax^3+bx^2+cx+d=0$? How do you tell whether $ax^3+bx^2+cx+d=0$ doesn't have any rational roots?

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The rational root theorem always works, hence the name theorem. – Git Gud Apr 27 '14 at 20:29
It always works - the root you give is not rational! – Old John Apr 27 '14 at 20:29
Possibly there is a typo: if the constant term is $5$ vs. $-5$ then it has root $\,x=5,\,$ which is easily found by the Rational Root Test, since it implies that the only possible rational roots are $\,\pm1,\pm 5.\ \$ – Bill Dubuque Apr 27 '14 at 20:42
See cubic formula. – Lucian Apr 28 '14 at 7:14

When you apply the rational root theorem, you find all the rational roots, if there are any. If the theorem finds no roots, the polynomial has no rational roots. (For a cubic, we would observe that the polynomial is irreducible over the rationals. This is because a factorization of the cubic is either the product of a linear factor and a quadratic factor or it is the product of three linear factors. Since in either case there is a linear factor, there would be a root in the rationals. With no rational root, we're done.)

Determining whether a polynomial has rational roots is done by means of the rational root test. There is no prior test one uses to determine whether to use the rational roots test; one starts with the test to find and remove all the "easy" factors. (One should also use the polynomial GCD and the derivative to remove repeated roots, but that's outside the scope of your question.) Only after all the easy roots are removed (or not found) do you move on to hard tools like the cubic formula.

For more discussion, see casus irreducibilis in the English Wikipedia.

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It is not true that no rational roots implies irreducible over $\,\Bbb Q,\,$ e.g. $\,(x^2+1)^2\$ – Bill Dubuque Apr 27 '14 at 20:36
@BillDubuque: Claim is true for cubics, the poster's question domain. Answer updated with this caveat. – Eric Towers Apr 27 '14 at 20:44
Glad to see the qualification added. Many students mistakenly believe the unqualified statement to be true. It would be even clearer to write "this implies" vs. "we would say", since the latter sounds like a definition, exactly the misconception we want to avoid. – Bill Dubuque Apr 27 '14 at 20:47
@BillDubuque: Yes, this is true. I've been reading (elementary) Galois theory recently and the possible factorings of the cubic is perhaps too clear to me presently. :-) – Eric Towers Apr 27 '14 at 20:48

The rational roots theorem yields potential rational roots. It says nothing about roots that are not rational. :)

To detect how many real roots there will be, take a look at the discriminant. If $a_1, a_2, a_3$ are roots of the polynomial, then:

$$\Delta = (a_1-a_2)^2(a_1-a_3)^2(a_2-a_3)^2 = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$$

Yuck! Note that $\Delta \geq 0 \iff$ there are $3$ real roots (counting multiplicity). Otherwise, there will be one real and two complex roots.

Actually finding these roots requires the God-awful cubic formula or certain numerical approximations, such as Newton's method, if you're fine with error.

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If you think the formula for the cubic is awful, check out the quartic.... – vonbrand Apr 28 '14 at 1:28
– Kaj Hansen Apr 28 '14 at 2:50