Decomposing polynomials with integer coefficients

Question

Can every quadratic with integer coefficients be written as a sum of two polynomials with integer roots? (Any constant $k \in \mathbb{Z}$, including $0$, is also allowed as a term for simplicity's sake.)

(In other words, is any given $P(x) = A + Bx + Cx^2$ expressible as

$$P(x) = \color{red}{k(x-r_1)(x-r_2)\cdots(x-r_n)} + \color{blue}{\ell(x-s_1)(x-s_2)\cdots(x-s_m)}$$

where all variables other than $x$ are integers?) As an example of such a decomposition, if $C = 1$ then $P(x) = (A - Ax) + (Ax + Bx + x^2) = \color{red}{-A(x-1)} + \color{blue}{(x)(x+A+B)}$. The "two polynomials" restriction is essential; expressions like $P(x) = \color{red}{(A)} + \color{green}{(Bx)} + \color{blue}{(Cx^2)}$ don't count.

I've been contemplating this statement for a while and could use some help. I'm having trouble whether trying to prove it or find a (verifiable) counterexample. (Note that the components can have arbitrarily high degrees $n,m$ but cancel out to give $P(x)$.) Variations on completing the square didn't help.

If the answer is affirmative, I would also be interested in the following generalizations:

• In addition to quadratics, can higher-order polynomials be decomposed into two polynomials?

• (Refinement of the above if it is true) If two polynomials do not suffice for $P(x)$ of arbitrary degree, is there a finite number $N$ that does?

Thanks in advance for any ideas or help.

^{Note: I have used the colors I can most easily distinguish in the question, but if they cause other people difficulty please feel free to change them or remove them.}

Answer 1

If $a$ is a multiple of $b$, so that $a =mb$, you can decompose in this way:
$$a+bx+cx^2=c(x-b)(x+b)+b(x+cb+m)$$
or, if you prefer:
$$a+bx+cx^2=cx^2+b(x+m)$$
But if $a$ can be expressed as $a=mb-n^2c$, where $m$ and $n$ are integers, then we can decompose the polynomial in this way: $$a+bx+cx^2=c(x-n)(x+n)+b(x+m)$$
Let's see an example that cannot be included in the previous cases ($a=mb$ and $a=mb-n^2c$).

Let $P(x)=2+57x+31x^2$, and let's decompose $P(x)$ in this way: $$2+57x+31x^2=k(x-r_1)(x-r_2)+\ell x+D \quad (1)$$ where $D=\ell s_1$, i.e. $D$ is a multiple of $\ell$.
From equation $(1)$ we can conclude: $$k=31$$ $$\ell = 57+31(r_1+r_2) \quad(2)$$ $$D=2-3(r_1 r_2)\quad(3)$$
But $D$ is a multiple of $\ell$ ($D=\ell s_1$), therefore we can conclude from that fact and from equations $(2)$ and $(3)$ the following: $$2-57s_1=31[r_1r_2+s_1(r_1+r_2)] \quad(4)$$ From equation $(4)$ we conclude that $2-57s_1$ is a multiple of $31$. Thus if $n=r_1 r_2+s_1(r_1+r_2)$ we reached the following equation: $$2=31n+57s_1 \quad(5)$$ A solution of equation $(5)$ is $n=-22$ and $s_1=12$ (Using Euclidean Algorithm).

Now we can solve equation $(4)$.
A solution is $r_1=-10$ and $r_2=49$ (Using factoring).
And replacing the values of $r_1$ and $r_2$ in $(2)$ we get: $$\ell = 1266$$ Therefore $P(x)=2+57x+31x^2$ can be decomposed in this way: $$2+57x+31x^2=31(x+10)(x-49)+1266(x+12)$$ A conclusion that we can draw from that example is:

If $a$ is not a multiple of $gcd(b,c)$, then the polynomial cannot be decomposed in this way: $$a+bx+cx^2=k(x-r_1)(x-r_2)+\ell(x-s_1)$$ since there won't be a solution to the Diophantine equation $(5)$.