I think I got the proof that no such real polynomial with degree $ \geq 6$ exists.
Let $n \geq 6$
Suppose for contradiction that $z_1,\ldots,z_n \in \mathbb R-\{0\}^n$ are such that $(X-z_1)...(X-z_n)=X^n+\sum_{k=1}^{n-1}z_iX^{n-i}$
Then three useful identities appear $$\sum_{k=1}^{n}z_k=-z_1 \; \; \; \;(1)$$
$$\sum_{\large1\leq i<j \leq n}z_iz_j=z_2 \; \; \; \;(2)$$
$$\prod_{k=1}^n z_k=(-1)^n z_n \; \; \; \;(3)$$
Since $$(\sum_{k=1}^{n}z_k)^2=\sum_{k=1}^{n}z_k^2+2\sum_{\large1\leq i<j \leq n}z_iz_j$$it follows that$$z_1^2=2z_2+\sum_{k=1}^{n}z_k^2$$
Hence $$0< \sum_{k=2}^{n}z_k^2=-2z_2 \; \; \; \;(4)$$ and $$0<\sum_{k=3}^{n}z_k^2=1-(z_2+1)^2 \; \; \; \;(5) $$
$(4)$ and $(5)$ imply $$\; \; \; \;-2<z_2<0 \; \; \; \;(6)$$
thus $(6)$ and $(4)$ imply $$0<\sum_{k=2}^{n}z_k^2 < 4 \; \; \; \; (7)$$
Also $(6)$ and $(5)$ imply $$0<\sum_{k=3}^{n}z_k^2 \leq 1 \; \; \; \; (8)$$
By AM-GM, $$\left(|z_3|^2\ldots|z_{n-1}|^2 \right)^{1/(n-3)} \leq \frac{1}{n-3}\sum_{k=3}^{n-1}z_k^2 \leq \frac{1}{n-3}\sum_{k=3}^{n}z_k^2$$
Hence
$$|z_3|^2\ldots|z_{n-1}|^2 \leq \left(\frac{1}{n-3}\sum_{k=3}^{n}z_k^2\right)^{n-3} $$
Squaring, $$|z_3|\ldots|z_{n-1}| \leq \left(\frac{1}{n-3}\sum_{k=3}^{n}z_k^2\right)^{\large \frac{n-3}{2}} \leq_{ \large (8)} \dfrac{1}{{(n-3)}^{(n-3)/2}} \; \; \; \; (9)$$
By triangle inequality $(1)$, and Cauchy-Schwarz
$$2|z_1| \leq \sum_{k=2}^{n}|z_k| \leq \sqrt{n-1} \sqrt{\sum_{k=2}^{n}z_k^2} $$
Hence by $(7)$,
$$|z_1| \leq \sqrt{n-1} \; \; \; \; (10)$$
Rewriting $(6)$ as $$|z_2|\lt2 \; \; \; \; (11) $$
Recalling $(3)$ (with $z_n$ cancelled from both sides) and putting together $(9)$, $(10)$ and $(11)$, we have
$$1=|z_1||z_2||z_3|\cdots|z_{n-1}| < \dfrac{ 2\sqrt{n-1}}{{(n-3)}^{(n-3)/2}}$$
This inequality fails for $n\geq 6$.
Contradiction.
I can't prove anything for $n=5$ so maybe the conjecture doesn't hold.
I'll give examples, then you'll understand the general pattern better.
Take a degree four polynomial. I'll denote a general fourth-degree polynomial by $P_4$. A general $5^{\text{th}}$ degree polynomial as $P_5$ and so on...
Your $P_4$ is: $ a_0x^4 + a_1x^3 + a_2x^2 + a_3x^1 + a_4$
Then, I'll get $4$ formulas:
$r_0 + r_1 + r_2 + r_3 = -\dfrac{a_1}{a_0}$
$r_0r_1 + r_0r_2 + r_0r_3 + r_1r_2 + r_1r_3 + r_2r_3 = +\dfrac{a_2}{a_0}$
$r_0r_1r_2 + r_0r_1r_3 + r_0r_2r_3 + r_1r_2r_3= -\dfrac{a_3}{a_0}$
$ r_0r_1r_2r_3 = + \dfrac{a_4}{a_0}$
The idea is, first you add products of single roots. That is, the first formula. Then, you add products containing 2 roots - second formula. Then you add products containing 3 roots - third formula. Then add products of 4 roots - fourth formula. And on the RHS, the denominator is always the coeffecient of the heighest power, and the signs alternate. The first formula always has the negative sign. The numerator is the coefficient matching with the numbers of roots in the multiplication, i. e. for product containing 4 roots, numerator should be A4.
Best Answer
There are other techniques, namely the polynomial long division as mentioned in a comment by pedja, the Ruffini's rule and for polynomials with integer coefficients the rational root theorem, but you can apply the method of equating coefficients to
$$\begin{eqnarray*} P(x) &:&=x^{4}-7x^{3}+3x^{2}+31x+20 =\left( x+1\right) \left( \alpha x^{3}+\beta x^{2}+\gamma x+\delta \right). \end{eqnarray*}\;\; \tag{1}$$
Expanding the RHS and comparing with the LHS, where the coefficient of $x^{4}$ is $1$, you would conclude that $\alpha =1$ and are left with a simpler equality
$$\begin{eqnarray*} P(x) &:&=x^{4}-7x^{3}+3x^{2}+31x+20=\left( x+1\right) \left( x^{3}+\beta x^{2}+\gamma x+\delta \right) \\ &=&x^{4}+\left( \beta +1\right) x^{3}+\left( \beta +\gamma \right) x^{2}+\left( \gamma +\delta \right) x+\delta . \end{eqnarray*}$$ Equating coefficients you get the simple system of four linear equations $$ \begin{equation*} \left\{ \begin{array}{c} \delta =20 \\ \gamma +\delta =31 \\ \beta +\gamma =3 \\ \beta +1=-7, \end{array} \right. \end{equation*}$$ whose solution is $\delta =20,\beta =-8,\gamma =11$. So
$$\begin{equation*} x^{4}-7x^{3}+3x^{2}+31x+20=\left( x+1\right) \left( x^{3}-8x^{2}+11x+20\right). \end{equation*}\tag{2}$$ Now by direct inspection (or by the rational root theorem, since $1,2,4$ and $5$ are the positive divisors of $20$) you can find that $x=-1$ is a zero of the cubic polynomial $Q(x):=x^{3}-8x^{2}+11x+20$, ie $Q(-1)=0$. Applying a similar method to $Q(x)$ you would find that
$$\begin{equation*} Q(x)=x^{3}-8x^{2}+11x+20=\left( x+1\right) \left( x-4\right) \left( x-5\right) . \end{equation*}$$
Consequently, $$\begin{equation*} x^{4}-7x^{3}+3x^{2}+31x+20=\left( x+1\right) ^{2}\left( x-4\right) \left( x-5\right). \end{equation*}\tag{3}$$
A shorter method is the Ruffini's rule applied to the polynomial division of $P(x)$ by $(x-r)$. This case (with $P(x)=x^{4}-7x^{3}+3x^{2}+31x+20$ and $r=-1,P(r)=0$, ) is shown bellow, where the remainder is $P(r)=s$.