Is this True? Question about polynomials degree and their product 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?
 A: $p \left(x \right)$ 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
A: 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.
A: 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
$\rho \in \Bbb C \setminus \Bbb R \tag 2$
satisfies
$p(\rho) = 0, \tag 3$
then
$p(\bar \rho) = 0 \tag 4$
as well; then
$x^2 - 2\Re(\rho) + \rho \bar \rho = (x - \rho)(x - \bar \rho) \mid p(x); \tag 5$
then
$(x^2 - 2\Re(\rho) + \rho \bar \rho) q(x) = p(x) \tag 6$
for some
$q(x) \in \Bbb R(x), \; \deg q(x) = 2, \tag 7$
since
$\deg q(x) + \deg (x^2 - 2\Re(\rho) + \rho \bar \rho) = \deg p(x) = 4; \tag 8$
on the other hand, if
$\exists \alpha \in \Bbb R, \; p(\alpha) = 0, \tag 9$
then
$p(x) = (x - \alpha) q(x), \; \deg q(x) = 3; \tag{10}$
since $\deg q(x)$ is odd, 
$\exists \beta \in \Bbb R, \; q(\beta) = 0; \tag{11}$
thus
$q(x) = (x - \beta)r(x), \; r(x) \in \Bbb R[x], \; \deg r(x) = 2; \tag{12}$
in this case
$p(x) = (x - \alpha)(x - \beta)r(x), \tag{13}$
and that does it.
