Proving real polynomials of degree greater than or equal to 3 are reducible this is the proof I'm given:

Example 1.20. Let $f \in \mathbb{R}[x]$ and suppose that $\deg(f) > 3$. Then $f$ is reducible.
Proof. By the Fundamental theorem of algebra there are $A \in C$ such that $$f (x) = (x - \lambda_1) \dots (x - \lambda_n).$$
Note that $0 = \overline{f(\lambda_j)} = f(\bar\lambda_j)$ since the coefficients of $f$ are real. Thus if $\lambda_j \in \mathbb{C} \setminus \mathbb{R}$ there is $k$ such that $\lambda_k = \bar\lambda_j$. Moreover $$(x - \lambda_j)(x - \bar\lambda_j) = x^2 + (\lambda_j + \bar\lambda_j) x + \lambda_j \bar\lambda_j = x^2 - 2 \Re(\lambda_j) + |\lambda_j|^2 \in \mathbb{R}[x].$$ Thus $f$ factorises into real polynomials of degree $1$ (corresponding to $\lambda_j \in \mathbb R$) and 2 (corresponding to a pair $\lambda_j, \bar\lambda_j \in \mathbb C$). ❑

What I don't understand is the step "$0 = \overline{f(\lambda_j)} = f(\overline{\lambda_j})$ since the coefficients of $f$ are real."
Firstly, why do we need this step for the rest of the proof?
Secondly, I don't follow how $f$ having real coefficients gives $0 = \overline{f(\lambda_j)} = f(\overline{\lambda_j})$.
Thanks

For every 2 complex numbers $a, b \in \mathbb{C}$ it's true that
$$\overline{a+b} = \overline{a} + \overline{b}$$
$$\overline{ab} = \overline{a} \overline{b}$$
And of course $a = \overline{a}$ for every $a \in \mathbb{R}$. So if $f(x) \in \mathbb{R}[x]$, then $\overline{f(x)} = f(\overline x)$.
