Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a polynomial function of degree $n \geq 1$. Let $x_1$ be a root of $f$ with $x_1 \in \mathbb{R}$. Show that there exists a unique function $q : \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x) = q(x)(x - x_1)$. Show that $q$ is a polynomial function with degree $n-1$.
I was thinking about using the intermediate value theorem. Because $f$ is a polynomial function, it is a continuous function on $\mathbb{R}$, so it is also continuous on interval $[x_1, x]$ for every $x > x_1$ (and continuous on $[x, x_1]$ for every $x < x_1$). Polynomials are also differentiable on $\mathbb{R}$, so it is differentiable on $(x_1 , x)$. Applying the intermediate theorem, there exists a $c \in [x_1, x]$ such that
$$\frac{f(x) - f(x_1)}{x-x_1} = f'(c).$$ Since $f(x_1) = 0$ we find $$f'(c) = \frac{f(x)}{x-x_1}.$$
If we apply it for the interval $[x,x_1]$ we find that there exists a $c_1 \in [x,x_1]$ such that $$\frac{- f(x_1)}{x_1-x} = f'(c_1).$$
I am just not sure how exactly I have to define $q(x)$ and what else I need to write to finish the proof.
We know we can write $$f(x)=q(x)(x-x_1)+r(x)$$ via normal polynomial division with $\deg r<\deg x-x_1=1$ (there are plenty of proofs for this fact online). But then plugging in $x_1$ gives us $$f(x_1)=q(x_1)(x_1-x_1)+r(x_1)\quad \implies \quad 0=r(x_1)$$ since $x_1$ is a root of $f$. However, $r$ is a constant since its degree is less than $1$, so $r(x_1)=0$, giving us $$f(x)=q(x)(x-x_1)$$ Now, we easily see that since degrees of polynomials add through multiplication, the degree of $q(x)$ must be $n-1$ since $f$ has degree $n$ and $x-x_1$ has degree $1$.
Uniqueness comes from the division algorithm.