Introduction

A detailed answer was given by Peter von Elbing. I’ll like to add to that and the other answers that the rational root theorem can also be used to find irreducible polynomial factors (here and in the following “irreducible” means irreducible over [math]\mathbb{Q}[/math] ) of higher than first degree. So, if the rational root theorem fails to find rational solutions, then that means that the polynomial has no linear factors of the form [math]q x - p[/math] with [math]p[/math] and [math]q[/math] integers that are relatively prime.

If the polynomial is of degree [math]4[/math] or higher and it doesn't have any rational roots, then it may be irreducible, or else it will factor into irreducible polynomials of degree 2 or higher. It then turns out that you can actually find such polynomial factors also using the rational theorem. I’ll give a detailed derivation of how this works for the case of factoring quartic polynomials into irreducible quadratic factors.

Factoring quartic polynomials into irreducible quadratic polynomials

We’re going to consider quartic polynomials with integer coefficients:

[math]\displaystyle x^4 + ax^3 + b x^2 + c x + d\tag{1}[/math]

If a polynomial is not monic (i.e. if the coefficient of the leading power of [math]x[/math], in this case [math]x^4[/math] is not equal to [math]1[/math]), then you make it monic by putting [math]x[/math] equal to [math]t[/math] divided by the coefficient of [math]x^4[/math], and then you multiply the resulting expression by the third power of the coefficient of [math]x^4[/math] in the original polynomial.

We then start with apply the rational root theorem to see if there are rational roots. If a rational root is found then we’re done because dividing the polynomial by [math]x[/math] minus the root then yields a [math]3[/math]rd degree polynomial and this has either [math]3[/math] rational roots, or one rational root and an irreducible quadratic factor (which we can find by dividing by [math]x[/math] minus that rational root), or no rational roots, in which case the third-degree polynomial is irreducible.

If there are no rational roots, we check if the polynomial is the square of a quadratic. If that’s the case, then [math]d[/math] must be a square. So, we check if [math]d[/math] is a square, if so, we compute the greatest common divisor of the polynomial with its derivative. If we find a greatest common divisor that’s not equal to [math]1[/math], then that can only be a quadratic polynomial and our quartic polynomial is then the square of the quadratic polynomial.

If [math]d[/math] is not a square or in case [math]d[/math] is a square but we've found that the quartic is not the square of a quadratic polynomial, then we want to find out if the polynomial factors into two different irreducible quadratic factors. We can then reduce the polynomial modulo [math]x^2 + p x + q[/math] for general integers [math]p[/math] and [math]q[/math]. We then demand that the resulting linear expression on [math]x[/math] is identical to zero, so we then set the coefficient of [math]x[/math] to zero as well as the constant term. This then gives us two equations for the two unknowns [math]p[/math] and [math]q[/math]. Reducing the polynomial modulo [math]x^2 + p x + q[/math] amounts to replacing [math]x^4[/math] by [math]-p x^3 - q x^2[/math], then replacing [math]x^3[/math] by [math]-p x^2 - q x[/math], and then replacing [math]x^2[/math] by [math]-p x - q[/math]. If we apply these steps to (1), we get:

[math]\displaystyle \left[p^2 (a - p) - p (b - q)- q (a - p) + c\right] x + p q (a - p) - q (b - q) + d[/math]

Equating the coefficient of [math]x[/math] to [math]0[/math] yields:

[math]\displaystyle \left(p^2 - q\right) (a - p) - p (b - q) + c = 0\tag{2}[/math]

Equating the constant term to zero yields:

[math]\displaystyle p q (a - p) - q (b - q) + d = 0\tag{3}[/math]

Multiplying (3) by [math]\frac{p}{q}[/math] and subtracting that from (2) yields:

[math]\displaystyle (q - \frac{d}{q})p - a q + c = 0\tag{4}[/math]

We must then distinguish between the case [math]q^2 = d[/math] and [math]q^2 \neq d[/math]. In the former case, both quadratic factors have the same value for [math]q[/math] which according to (4) will be equal to [math]\frac{c}{a}[/math]. We can then test if this is a possibility, by testing if [math]d[/math] is a square and of that’s the case then to test if [math]d = \frac{c^2}{a^2}[/math]. If this is the case, then we can substitute [math]q = \frac{c}{a}[/math] in (3). This then yields a quadratic equation for [math]p[/math]:

[math]\displaystyle a p^2 - a^2 p + a b - 2 c = 0\tag{5}[/math]

where for [math]d[/math] we substituted [math]\frac{c^2}{a^2}[/math]. This equation must then have two integer solution. If not, then either the polynomial is irreducible, or it does factor into two quadratics but the two values for [math]q[/math] are different. To test the latter possibility, we then proceed with the case [math]q^2 \neq d[/math].

If [math]q^2 \neq d[/math], then we use (4) to express [math]p[/math] in terms of [math]q[/math]:

[math]\displaystyle p = \frac{c-a q}{d - q^2} q\tag{6}[/math]

Substituting this in (3) and expanding out the equation then yields:

[math]\displaystyle \begin{split} &q^6 - b q^5 + (a c - d) q^4 + (2 b d - c^2 - a^2 d) q^3\\ & + (a c d - d^2)q^2 - b d^2 q + d^3 = 0 \end{split}\tag{7}[/math]

So, we see that [math]q[/math] satisfies a 6th degree polynomial equation. But we know that [math]q[/math] must be an integer if it’s true that the polynomial does indeed factor into two quadratics. This means that the rational root theorem applies, so we can very easily determine the integer solutions for [math]q[/math]. If there are integer solutions, then we need to compute the corresponding values for [math]p[/math] using (6). Any pair of [math]q[/math] and the corresponding [math]p[/math] that are both integers will then define a quadratic factor [math]x^2 + p x + q[/math] of the polynomial.

If we arrived at (7) in case of [math]d = \frac{c^2}{a^2}[/math] after (5) yielded no integer solutions for [math]p[/math], then (7) will have a double root at [math]q = \frac{c}{a}[/math], so we can then factor out [math]\left(q - \frac{c}{a}\right)^2[/math]from the polynomial. We then get the 4th degree polynomial equation:

[math]\displaystyle q^4 + (2 r - b) q^3 + (a c - 2 b r + 2 r^2)q^2 + (2 r - b) r^2 q - r^4=0[/math]

where [math]r = \frac{c}{a}[/math] for which we must then seek integer solutions for which the corresponding values of [math]p[/math] that follow from (6) are also integers.

Example

Using Mathematica I expanded out the product of two monic quadratic polynomials with random integer coefficients between [math]-200[/math] and [math]200[/math], such that I could only see the end result, I had no access to the quadratic polynomial factor. The generated quartic polynomial was:

[math]\displaystyle x^4-51 x^3-10746 x^2+20334 x-7192[/math]

Applying the rational root theorem does not yield any integer roots. The constant term [math]d[/math] in here is not a square which combined with the fact that there are no integer roots, means that we don’t need to compute the greatest common divisor of the polynomial with its derivative, as this will be 1. And we must skip the tests that deal with the possibility that the two values for [math]q[/math] of the constant terms of the quadratic factors are the same.

We then proceed with computing the 6th degree polynomial in (7) for this case. This yields:

[math]\displaystyle \begin{split} R(x) &= q^6+10746 q^5-1029842 q^4-240194700 q^3\\ &+7406623664 q^2+555835388544 q-372005221888\end{split}\tag{8}[/math]

We then seek integer solutions for this for which [math]p[/math] given by (6) as

[math]\displaystyle p = -\frac{ 51 q+20334}{q^2 + 7192}q\tag{9}[/math]

is also an integer. The constant term [math]R(0)[/math] in (8) is [math]-372005221888[/math] and this is easily factored as:

[math]\displaystyle 372005221888 = 2^9\times 29^3 \times 31^3tag{10}[/math]

Checking if [math]q = 1[/math] is a root yields:

[math]\displaystyle R(1) = 190995576525[/math]

We then apply the rational root theorem to the polynomial [math]R(1+x)[/math]. The possible integer roots of [math]R(1+x)[/math] are the divisors of the constant term, which is [math]R(1)[/math]. So, we don't need to expand out [math]R(1+x)[/math]. Since adding [math]1[/math] to any possible root of [math]R(1+x)[/math] yields possible roots of [math]R(x)[/math] we can then shrink the list of possible roots a lot. Factoring [math]R(1)[/math] yields:

[math]\displaystyle 190995576525= 3^2\times 5^2\times 13\times 61\times 1070453 \tag{11}[/math]

If we compare (11) to (10), we see that[math] 61 + 1 = 2\times 31[/math], so [math]62[/math] will be a common entry of the lists of integer root candidates implied by applying the rational root theorem to [math]R(x)[/math] and [math]R(1+x)[/math]. According to (9) the value for [math]p[/math] for [math]q = 62[/math] is[math] p = -132[/math]. Since the value of p is an integer, we have fond the quadratic factor [math]x^2 + p x + q = x^2 - 132 x + 62[/math]. Dividing the polynomial by this quadratic factor then yields the complete factorization as:

[math]\displaystyle \begin{split} & x^4-51 x^3-10746 x^2+20334 x-7192\\ & = \left( x^2 - 132 x + 62\right)\left( x^2 +81 x -116\right)\end{split}[/math]

Note that instead of dividing the quartic by the quadric factor we found, we could also have divided [math]d[/math] by [math]q[/math] to find the other value for [math]q[/math] for the other quadratic factor, and inserting that in (9) will then yield the corresponding value for [math]p[/math] for that other quadratic factor.