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Let $f \in \mathbb R [x]$ and suppose that $\deg(f) \geq 3$. Then $f$ is reducible.

Proof: By the Fundamental theorem of algebra there are $\lambda _j \in \mathbb C$ such that $$f(x) = (x-\lambda_1) \dots (x-\lambda_n) .$$ Note that $0 = \overline {f(\lambda_j)} = f( \overline {\lambda _j} )$ since the coefficients of $f$ are real. Thus if $\lambda _j \in \mathbb C \setminus \mathbb R$ there is $k$ such that $\lambda _k = \overline \lambda _j$. Moreover $$(x-\lambda _j) (x- \overline {\lambda _j}) = x^2 -(\lambda_j + \overline {\lambda _j}) + \lambda _j \overline {\lambda _j} = x^2 -2 \rm {Re} (\lambda) x + |\lambda|^2 \in \mathbb R [x] .$$ Thus $f$ factorises into real polynomials of degree $1$ (corresponding to $\lambda_j \in \mathbb R$) and $2$ (corresponding to a pair $\lambda_j, \overline {\lambda_j} \in \mathbb C \setminus \mathbb R$).

I have a few queries about this proof.

1). Why is the author considering $\lambda_j \in \mathbb{C} \backslash \mathbb{R}$ and claiming that $\lambda_k = \bar{\lambda_j}$. Is $\lambda_k$ real here?

2). If the degree of $f$ is more than or equal to $3$, then why does he go on to consider a quadratic?

3) I don't really understand the conclusion from the above lines.

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Whole point is that if $\lambda_j$ is a root then $\bar{\lambda}_j$ is also a root. By the choice of $\lambda_j$, $\lambda_k$ is not real. And the quadratic he considers is something which divides the original polynomial $f$. Ie the quadratic divides $f$ and has degree strictly less than $f$ so $f$ is reducible by definition.

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