I was pondering the existence of cubic polynomials of the form
$$ax^3 + bx^2 + cx + d,$$
where the constants $a$, $b$, $c$ and $d$ are all non-zero integer consecutive terms of an arithmetic progression, such that the polynomial can be factorized as a product of three linear factors of the form $(ex+f)(gx+h)(ix+j)$, where $e,f,g,h,i$ and $j$ are integers.
Can anyone think of an example of such a polynomial? Is there a general rule to describe them?
Best Answer
If you plot a bunch of these polynomials you'll notice that they all only have one real root, so they don't admit such a factorization even over the reals. This turns out to be true! Here's a somewhat nasty proof but it works. Write the polynomial as
$$p(x) = (a - 3d) x^3 + (a - d) x^2 + (a + d) x + (a + 3d)$$
(an arithmetic progression but where $a$ is the average instead of the initial term and the common difference is $2d$). WolframAlpha very helpfully tells us that the discriminant of this polynomial is
$$-16 (a^4 - 22 a^2 d^2 + 125 d^4) = -16 ((a^2 - 11d^2)^2 + 4d^4)$$
and in particular it is always negative, which means $p(x)$ has exactly one real root.
(Maybe Sturm's theorem or Budan's theorem could be used to give a different proof?)