Consider a polynomial $p\in\mathcal{P}(\mathbb{R})$ with a complex root $\lambda$.
Since the coefficients of $p$ are real, we know that complex roots occur in conjugate pairs.
We can write $p(x)$ as
$$p(x)=(x-\lambda)(x-\bar{\lambda})q(x)$$
$$=(x^2-2\text{Re}(\lambda)x+|\lambda|^2)q(x)$$
for some polynomial $q\in\mathcal{P}(\mathbb{C})$ with degree two less than the degree of $p$.
Suppose we prove that $q$ has real coefficients.
My question is what is the induction argument that proves that in the factorization of $p$ into its root factors we always get the same number of $x-\lambda$ and $x-\bar{\lambda}$ factors.
Here is the induction argument I came up with.
The degree of $p$ must be $\geq 2$ since we already have two roots $\lambda$ and $\bar{\lambda}$.
$p$ with degree 2 is the base case. It can be factored as $(x-\lambda)(x-\bar{\lambda})$ so the factors with $\lambda$ and $\bar{\lambda}$ occur the same number of times.
Now suppose as inductive hypothesis that for all $n\in\mathbb{N}$ such that $n\leq N\in\mathbb{N}$ we have that if $\deg{p}=n$ then $x-\lambda$ and $x-\bar{\lambda}$ occur the same number of times in $p$'s factorization.
Suppose $\deg{p}=N+1$.
Then $\deg{q}=N-1$.
Since $p(x)=(x-\lambda)(x-\bar{\lambda})q(x)$ then it is clear that $x-\lambda$ and $x-\bar{\lambda}$ occur the same number of times (once more each than they appear in $q(x)$).
By induction we infer that for all $n\in\mathbb{N}$ such that $n\geq 2$, if $p$ is a polynomial with real coefficients of degree $n$ and if $\lambda$ is a complex root then the factors $(x-\lambda)$ and $(x-\bar{\lambda})$ appear in the factorization of $p$ the exact same number of times.
Is this argument correct?
You need to be more careful in the induction step: the polynomial $p$ may for example not have a complex root at all. You could distinguish between the cases "$p$ has a real root" and "$p$ has only complex roots".