Non- zero polynomial can be written as product of polynomials of degree at most two 
Prove that every non-zero polynomial $\in \mathbb{R}[x]$ can be written as product of polynomials in $\mathbb{R}[x]$ of degree at most two.

Is there elementary proof, at high school level ? 
Please, suggest me on how to prove or lead me to the link.Thank you. 
(Fundamental theorem of Algebra is quite hard for me.)
 A: We need to assume that if $n$ is a degree of our polynomial $f$ then $f$ has $n$ roots from $\mathbb C$.
Try to prove the following statements.

*

*

If $z\in\mathbb C$ is a root of our polynomial then $\bar{z}$ is a root of $f$.



*

If previous  $z=a+bi$, where $a$ and $b$ are reals, then $f$ divided by $x^2-2ax+a^2+b^2$.



*

If $c\in\mathbb R$ is a root of $f$ thus, $f$ divided by $x-c$

and by induction we are done!
A: I will prove the statement by strong induction. Of course, the statement is trivial if the degree of the polynomial is $1$ or $2$. Suppose that you have a polynomial $P(x)\in\mathbb{R}[x]$ of degree $n>2$ and assume that every non-constant polynomial in $\mathbb{R}[x]$ whose degree is smaller than $n$ can be written as a product of polynomials with degree $1$ or $2$.
Let $x_0$ be a root of $P(x)$; the Fundamental theorem of Algebra assures that such a root existes. Let us consider two possibilites: $x_0\in\mathbb R$ and $x_0\notin\mathbb R$.


*

*Since $x_0\in\mathbb R$ and $x_0$ is a root of $P(x)$, $P(x)$ can be written as $Q(x)(x-x_0)$, with $Q(x)\in\mathbb{R}[x]$. Since the degree of $Q(x)$ is $n-1$, $Q(x)$ can be written as product of polynomials whose degree is $1$ or $2$ and therefore $P(x)$ has the same property.

*It happens that $\overline{x_0}$ is a root of $P(x)$ too, because, since the coefficients of $P(x)$ are real, $\overline{P\bigl(\overline{x_0}\bigr)}=P(x_0)=0$ and therefore $P\bigl(\overline{x_0}\bigr)=0$. Since $x_0$ and $\overline{x_0}$ are distinct roots of $P(x)$, $P(x)$ can be written as $Q(x)(x-x_0)\bigl(x-\overline{x_0}\bigr)$. Furthermore,$$(x-x_0)\bigl(x-\overline{x_0}\bigr)=x^2-2\operatorname{Re}(x_0)x+|x_0|^2\in\mathbb{R}[x]$$and this implies that $Q(x)\in\mathbb{R}[x]$. Since the degree of $Q(x)$ is $n-2$, $Q(x)$ can be written as product of polynomials whose degree is $1$ or $2$ and therefore $P(x)$ has the same property.

