The statement is

Every degree $4$ polynomial with real coefficients is expressible as the product of two degree $2$ polynomials with real coefficients.

This and much more general versions are of course simple consequences of the Fundamental Theorem of Algebra and there are of course numerous classical ways to even calculate all the roots of the quartic, thereby establishing much stronger statements constructively.

However, suppose that we are only interested in proving the existence of such a factorisation and do not care about what the quadratics are. What's more, we want to do this for quartics only and there is no need for the argument to be applicable to polynomials of higher even degree.

Do we still need an instance of the FTA and or a classical method that does this by working out all the coefficients? Or is there now a more elementary (perhaps purely existential) argument? Perhaps one justifying the expressibility of the quartic as the difference of two polynomials squared?

This question is inspired by a similar question asked by a high school student and I am actually wondering if there is a proper proof of the above statement that is within the grasp of a high school student, i.e. no FTA, in fact no complex numbers at all, and no extensive calculations of roots/coefficients (and setting up four nonlinear equations simply by comparing coefficients falls into this class).

  • $\begingroup$ Since the degree 4 poly. will have 4 non-zero real roots it could be written as $f(x)=[(x-r_1)(x-r_2)][(x-r_3)(x-r_4)[$ where each square bracket holds a degree 2 polynomial. What do you see wrong with this? $\endgroup$ – NoChance May 12 at 5:28
  • 1
    $\begingroup$ It does not need to have real roots, e.g. $x^4+1$. $\endgroup$ – Damian Reding May 12 at 10:02
  • $\begingroup$ Still there are 4 roots and my comment applies. $\endgroup$ – NoChance May 12 at 11:45
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    $\begingroup$ How do you justify that there are roots at all? $\endgroup$ – Damian Reding May 12 at 12:21
  • $\begingroup$ À propos of nothing, $x^4+1=(x^2+\sqrt2x+1)(x^2-\sqrt2x+1)$. $\endgroup$ – Lubin May 13 at 0:57

After a linear transformation, one may assume that the quartic has the form $x^4 + a x^2 + b x + c$. As suggested, let's try to write

$$x^4 + a x^2 + b x + c = (x^2 + t)^2 - (r x + s)^2$$

over $\mathbf{R}$, which will give the desired factorization. In order to achieve this, it suffices to show that we can choose $t$ so that the difference

$$\Delta:=x^4 + a x^2 + b x + c - (x^2 + t)^2$$

has the following properties:

  1. The leading coefficient of $x^2$ is negative.
  2. There are repeated roots.

This implies it has the form $-(rx + s)^2$. The leading coefficient of $\Delta$ is $a - 2t$, so for condition 1, we need $2t > a$. For condition 2, the discriminant of $\Delta$ factors as

$$(2t-a)(8 t^3 - 4 a t^2 - 8 c t + 4 a c - b^2) =:(2t-a)g(t)$$

To summarize, it suffices to show that the second factor $g(t)$ above has a real root $2t > a$. But one finds:

$$g(a/2) = -b^2, \quad g(\infty) = + \infty.$$

So, as long as $b \ne 0$, we deduce that then $g(t)$ has a real root $2t > a$ by the intermediate value theorem, and we are done.

This still leaves the case when $b = 0$ which one can do by hand directly. In this case, we have $$x^4 + a x^2 + c = (x^2 + a/2)^2 - (a^2/4 - c),$$ which works if $a^2/4 \ge c$ and hence in particular when $c$ is negative, or $$x^4 + a x^2 + c = (x^2 + \sqrt{c})^2 - (2 \sqrt{c} - a) x^2,$$ which works in the complementary case $c \ge a^2/4$.

  • $\begingroup$ I suppose you meant $-(rx+s)^2$, not $(-rx+s)^2$? $\endgroup$ – Hans Lundmark May 12 at 11:27
  • $\begingroup$ Typo corrected! $\endgroup$ – user670344 May 12 at 15:07
  • $\begingroup$ Nice! This is 100% to the point and exactly what I envisioned: the existence of t, r, s is proved without constructing them. Many thanks. $\endgroup$ – Damian Reding May 12 at 15:31

It's possible to reduce the problem to demonstrating that a particular 6th degree polynomial has 2 solutions. I have not found an easy way to do that, though. This works as follows.

Suppose $p(x) = x^4 + a x^3 + b x^2 + c x + d$ is a square-free quartic polynomial. We then want to find real numbers $r$ and $s$ such that the quadratic polynomial $d(x) = x^2 - r x - s$ is a factor of $p(x)$. If we reduce $p(x)$ modulo $d(x)$ then we end up with a linear function which should be identically zero if $r$ and $s$ are chosen such that $d(x)$ divides $p(x)$. This then yields polynomial equations for $r$ and $s$, so the task is reduced to demonstrating that these equations are guaranteed to have real solutions.

A simple computation yields:

$$p(x) \bmod d(x) = a r s+b s+d+r^2 s+s^2 + \left(c + b r + a r^2 + r^3 + a s + 2 r s\right)x$$

Demanding that this is zero for all $x$ implies that the constant term and the coefficient of $x$ is zero. Since $p(x)$ is square-free, if two quadratic factors have the same value for $r$, the $s$ must be different for those factors, or vice versa. To deal with the former case, we can eliminate $s$ using the two equations. In that case we find:

$$r = \frac{a s^2+c s}{d-s^2}$$


$$s^6 + b s^5 + (ac-d)s^4 + \left(a^2 d-2 b d+c^2\right)s^3 + \left(a c d-d^2\right)s^2 + bd^2 s + d^3 = 0$$

We end up with a 6th degree equation because there are $\binom{4}{2} = 6$ ways to choose 2 roots from the 4 roots of the quartic to form a quadratic factor. We can simplify the above equation, by exploiting the symmetry $s\rightarrow \dfrac{d}{s}$ that derives from the fact that multiplying the values of the two quadratics that divide the quartic must yield $d$. If we put $s + \dfrac{d}{s} = u$ we get the equation:

$$u^3 + b u^2 + (a c - 4 d) u + a^2 + c^2 - 4 b d = 0$$

In the other case, we have:

$$s = -\frac{d (a+2 r)}{r^3 + 2 a r^2 +\left(b+a^2\right)r + ab-c}$$


$$ \begin{split} r^6 &+ 3a r^5 + \left( 3a^2 + 2 b\right)r^4 + \left( a^3 + 4 a b\right)r^3 + \left(2 a^2 b+a c+b^2-4 d\right) r^2\\ & + \left(a^2 c+a b^2-4 a d\right) r +a b c-c^2 -a^2 d = 0 \end{split} $$

Because this remains invariant under $r\rightarrow -\dfrac{a}{2} + r$, we can substitute $\left(r+ \dfrac{a}{2}\right)^2 = u$ to get to the third degree equation:

$$ \begin{split} u^3 &+ \left(8 b^2 - 3 a^2\right)u^2 + \left(3 a^4 -16 a^2 b +16 b^2 + 16 a c -64 d\right)u\\ &+ 64 a b c + 8 a^4 b - 16 a^2 b^2- 16 a^3 c -64 c^2 - a^2=0 \end{split}$$


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