Is It Possible to Have Quartic Equation that Has Always Rational Root?

Question

Let take the quartic equation $ax^4+bx^3+cx^2+dx+e=0$, how can we find/construct $a, b, c, d,e$ such that root $x$ is always a rational number? i.e. a general condition on $a, b, c, d, e$ such that all the roots of the equation are rational.

We can use rational root theorem but that depends on root, I would like to find independent of the root, I don't need to have the root but the confirmation that the root is always rational.

For example it might look like $ax^4+(b+2a)x^3+(c+d)/2 \times x^2+dx+e$, in general, $a, b, c, d,e$ could be expressed in some rational variables and be true for any arbitrary values.

Answer 1

Let

$r_1, r_2, r_3, r_4 \in \Bbb Q; \tag 1$

then set

$f(x) = (x - r_1)(x - r_2)(x - r_3)(x - r_4)$ $= x^4 - (r_1 + r_2 + r_3 + r_4)x^3 + (r_1 r_2 + r_1 r_3 + r_1 r_4 + r_2 r_3 + r_2 r_4 + r_3 r_4)x^2$ $- (r_1 r_2 r_3 + r_1 r_2 r_4 + r_2 r_3 r_4) x + r_1 r_2 r_3 r_4; \tag 2$

finally, for any

$\rho \in \Bbb R \tag 3$

we take

$a = \rho, \tag 4$

$b = -\rho(r_1 + r_2 + r_3 + r_4), \tag 5$

$c = \rho(r_1 r_2 + r_1 r_3 + r_1 r_4 + r_2 r_3 + r_2 r_4 + r_3 r_4), \tag 6$

$d = -\rho(r_1 r_2 r_3 + r_1 r_2 r_4 + r_2 r_3 r_4), \tag 7$

$e = \rho r_1 r_2 r_3 r_4; \tag 8$

then the quartic polynmial

$p(x) = \rho f(x) = ax^4 + bx^3 + cx^2 + dx + e \in \Bbb R[x] \tag 9$

has zeroes preciely as in (1). Indeed, every real quartic with roots (1) takes this form, since it must be divisible by $f(x)$ with quotient of degree 0, that is, a real number.

Answer 2

Well, there are formulas for finding the roots of a general quartic $ax^4 + bx^3 + cx^2 + dx + e$ in this link; you can play with those to find some conditions to ensure the quartic will have some rational root.

Answer 3

IMO, the easiest way to generate those polynomials is to choose four rationals (in increasing order $r_1< r_2< r_3< r_4$ for uniqueness), and to expand

$$(x-r_1)(x-r_2)(x-r_3)(x-r_4).$$