It seems to be true that every real polynomial $p_n$ of degree $n$ can be factored in the following (not unique) way $$ p_n = \sum_{i=0}^n a_ix^i = s \left\{\prod_{i=1}^n(x-b_i)\right\} +t \tag{$\ast$}$$ with $a_i, s, b_i, t $ all in $\mathbb{R}$.
For example:
$x-1=(x-1)+0$
$x^2-5x+7=(x-2)(x-3)+1=(x-5)(x-0)+7$
$x^2+1 = (x-0) \cdot (x-0) +1$
I am having a hard time coming up with a rigorous proof. I have tried using the fact that every polynomial $p_n$ can be written as $(x-a)g(x)+b$, where $g$ is a polynomial function and $a$ is an arbitrary real number, but with no success. Another way it could be done is by expanding the right-hand side of $(\ast)$ and showing that the linear system of equations with the coefficients of matching powers of $x$ has always a real solution. But I feel there must be a more simple proof out there, e.g. by induction, if possible without the Fundamental Theorem of Algebra. Would be glad if someone could enlighten me!
No, this is not possible in general. For instance, let $p(x)=x^3+x$. Note that $p'(x)=3x^2+1$ is always positive, so $p$ is strictly increasing. That means that for any $t\in\mathbb{R}$, $p(x)-t$ has at most one real root. Moreover, $p(x)-t$ cannot have any repeated real roots, since its derivative is never $0$. So there is no $t$ such that $p(x)-t$ can be factored into linear factors over $\mathbb{R}$.
In fact, you can prove this without any calculus as follows. First, note that $x^3+x=x(x^2+1)$ is strictly increasing: when $x>0$ it is obviously positive and increasing, and when $x<0$ it is negative and increasing since both factors are decreasing in absolute value. So all that remains is ruling out the possibility that $x^3+x-t$ has a single real triple root for some $t$. But if $b\in\mathbb{R}$ were a triple root of $x^3+x-t$, we would have $x^3+x-t=(x-b)^3=x^3-3bx^2+3b^2x-b^3$. Comparing the quadratic terms gives $b=0$ but then the linear terms do not match, so this is impossible.