Can *any* real polynomial be factored linearly (plus a constant) over $\mathbb{R}$? 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!
 A: 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.
A: Your claim is that after a suitable vertical translation, the graph of a degree $n$ real polynomial function will intersect the $x$-axis $n$-times, multiplicities counted. Or in other words, the polynomial will have only real roots. This is true in degrees $\leq 2$, but fails at $$P(x)=x^3+x.$$
