Is it true that every polynomial of degree 4 and real coefficients can be expressed as the product of two polynomials of degree two and of real coefficients? In case of an affirmatively answer prove it and in case of answering negatively give a counterexample.

I'm having a lot of struggle with this one:

Things I tried so far: ax⁴ + bx³ + cx² + dx + e and inventing a root to bring it down with ruffini but I could not get to a result.

Also I tried inventing a polynom which is: x⁴ - x³ + x² -x which i could bring down with ruffini twice with root = 0. So the polynomic left was x² - x + 1 that when multiplied by the other polynom which is (x)(x) = x² it got me to:

x⁴ - x³ + x²

So im really confused here, can someone help?

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    $\begingroup$ Here is a hint: any real polynomial of degree $n$ has $m$ complex roots and $r$ real roots where $n=m+r$. What can you say about $m$ in general (think complex conjugates)? Then use this information to do some case analysis. $\endgroup$ – Wraith1995 Jul 13 '18 at 4:53
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    $\begingroup$ Wow that is great, I can say to start that if a polynomial has real coefficients and let's say z1 is root of the polynomial, then the conjugate of z1 is also root. Im gonna keep working on this one, thank you for the hint $\endgroup$ – Juan Ignacio Bruzzone Jul 13 '18 at 4:59
  • $\begingroup$ That only works for non-real $z$, but yeah, basically. $\endgroup$ – Robert Lewis Jul 13 '18 at 5:00
  • $\begingroup$ The number of possible (real, complex) roots are (4,0), (2,2), or (0,4)$ and the complex roots occur in conjugate pairs. $\endgroup$ – steven gregory Jul 13 '18 at 5:04
  • $\begingroup$ So claiming that if r is root of P(x), then -r is root of P(x). And also that 2r is root of P(x), then -2r is root aswell. We can write the polynomial ax4 +bx3 +cx2 +dx +e as: a(x - r)(x + r)(x -2r)(x + 2r) = (ax(2) - ar(2)).(x(2) -4r(2)) Which results in: ax(4) -5ar(2)x(2) + 4ar(4) so we can claim that +5ar(2) is c and +4ar(4) is e. And all others are 0. We can prove this statement? $\endgroup$ – Juan Ignacio Bruzzone Jul 13 '18 at 5:08

Yes it's true. If

$p(x) \in \Bbb R[x], \; \deg p(x) = 4, \tag 1$

and $p(x)$ has no real roots, then if

$\rho \in \Bbb C \setminus \Bbb R \tag 2$


$p(\rho) = 0, \tag 3$


$p(\bar \rho) = 0 \tag 4$

as well; then

$x^2 - 2\Re(\rho) + \rho \bar \rho = (x - \rho)(x - \bar \rho) \mid p(x); \tag 5$


$(x^2 - 2\Re(\rho) + \rho \bar \rho) q(x) = p(x) \tag 6$

for some

$q(x) \in \Bbb R(x), \; \deg q(x) = 2, \tag 7$


$\deg q(x) + \deg (x^2 - 2\Re(\rho) + \rho \bar \rho) = \deg p(x) = 4; \tag 8$

on the other hand, if

$\exists \alpha \in \Bbb R, \; p(\alpha) = 0, \tag 9$


$p(x) = (x - \alpha) q(x), \; \deg q(x) = 3; \tag{10}$

since $\deg q(x)$ is odd,

$\exists \beta \in \Bbb R, \; q(\beta) = 0; \tag{11}$


$q(x) = (x - \beta)r(x), \; r(x) \in \Bbb R[x], \; \deg r(x) = 2; \tag{12}$

in this case

$p(x) = (x - \alpha)(x - \beta)r(x), \tag{13}$

and that does it.

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    $\begingroup$ Wow that was very complex, I don't think I'm at that level to solve this problem. Thank you a lot! $\endgroup$ – Juan Ignacio Bruzzone Jul 13 '18 at 5:23
  • $\begingroup$ @JuanIgnacioBruzzone: it might look complicated, but it's really not. The main thing is that if $\mu$ solves $f(x) = 0$, then $f(x) = (x - \mu)g(x)$, that and if $f(x)$ is real, and $\mu \in \Bbb C \setminus \Bbb R$, then $f(\bar \mu) = 0$ as well; then we divide things into two cases: either $f(x)$ has a complex root, or it does not . . . Cheers! $\endgroup$ – Robert Lewis Jul 13 '18 at 5:29

By the fundamental theorem of algebra, a complex polynomial of degree $n$ can be expressed as the product of complex linear factors. This, of course, applies for real polynomials, with the caveat that the linear factors are still complex.

It's simple to see that $\overline{zw} = \overline{z}\overline{w}$ and $\overline{z + w} = \overline{z} + \overline{w}$. From this we see that $\overline{z^k} = \overline{z}^k$, and hence $p(\overline{z}) = \overline{p(z)}$, where $p$ is a real polynomial. So, if $(z - (x + iy))$ is a factor of $p(z)$, then $p(x + iy) = 0$, hence $p(x - iy) = 0,$ and thus $(z - (x - iy))$ is a factor of $p(z)$ as well.

When $y = 0$, this is the same factor, but when $y \neq 0$, i.e. when $x + iy$ isn't real, then we obtain a factor of $$(z - (x + iy))(z - (x - iy)) = (z - x + iy)(z - x - iy) = (z - x)^2 - (iy)^2 = (z - x)^2 + y^2.$$

If $p(z)$ has no complex roots, then we can write $p(z)$ as a product of $4$ linear factors, which can be paired any way you see fit.

If $p(z)$ has no real roots, then there must be four complex roots in two conjugate pairs. As above, we can obtain two quadratic factors.

If $p(z)$ has at least one real and at least one complex root, then we obtain one quadratic factor as above, and multiply the two real linear factors to obtain the other quadratic factor.


$p \left(x \right)$ being a polynomial of degree 4. Its following possibilites for roots

Case-$1$ $$4 \text{ Real Roots-Answer is Trivial}$$

Case-$2$ $$2 \text{ Real Roots }-\alpha, \beta\text{ ; }2\text{ Complex Roots }-\gamma, \overline{\gamma}$$

$$ p \left(x \right)=\left(x-\alpha \right)\left(x-\beta \right)\left(x^2-(\gamma+\bar\gamma)x+\gamma\bar\gamma \right)$$

Case-$3$ $$4\text{ Complex Roots }-\alpha, \overline{\alpha},\beta, \overline{\beta}$$

$$ p \left(x \right)=\left( x^2-(\alpha+\bar\alpha)x+\alpha\bar\alpha\right)\left( x^2-(\beta+\bar\beta)x+\alpha\bar\beta \right)$$

Note that Complex Roots appear in pairs


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