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
 A: In general, if $w,z\in\Bbb C,$ then $$\overline{w+z}=\overline w+\overline z,$$ and $$\overline{wz}=\overline w\,\overline z.$$ From these, it follows that if $$f(x)=a_nx^n+\cdots+a_1x+a_0,$$ then $$\overline{f(x)}=\overline{a_n}\,\overline x^n+\cdots+\overline{a_1}\,\overline x+\overline{a_0}.$$ Furthermore, given $z\in\Bbb C,$ we have $\overline z=z$ if and only if $z\in\Bbb R,$ so if $f(x)$ has real coefficients, then $\overline{a_k}=a_k$ for $k=1,...,n,$ and so we have $$\overline{f(x)}=a_n\overline x^n+\cdots+a_1\overline x+a_0=f\left(\overline x\right).$$ Now, suppose that $\lambda$ is a root of $f,$ meaning $f(\lambda)=0.$ Then $$0=\overline0=\overline{f(\lambda)}=f\left(\overline\lambda\right),$$ and so $\overline\lambda$ is also a root of $f.$ If $\lambda$ is real, then $f(x)$ has the real factor $(x-\lambda),$ so is reducible. If $\lambda\in\Bbb C\setminus\Bbb R,$ then $f(x)$ has the factor $$(x-\lambda)\left(x-\overline\lambda\right)=x^2-\left(\lambda+\overline\lambda\right)x+\lambda\overline\lambda,$$ which can be shown to have real coefficients, and so $f(x)$ is reducible.
A: In general, if $f(x)$ has real coefficients, then $\overline {f(x)}=f(\overline x)$. This follows from the following fact. 
Consider the conjugate function
$$\begin{align} \phi: \ \ \ \mathbb{C} \ \ \ & \to \  \ \  \mathbb{C}\\ a+bi& \mapsto a-bi\end{align}$$
Note that $\phi(\lambda\gamma)=\phi(\lambda)\phi(\gamma)$, $\phi(\lambda+\gamma)=\phi(\lambda)+\phi(\gamma)$ for all $\lambda,\gamma \in \mathbb{C}$ and $\phi(\alpha)=\alpha$ for all $\alpha\in \mathbb{R}$ (just check it by yourself), then it follows that, if $f(x)=a_0+a_1x+\cdots a_nx^n$ where $a_0,...,a_n$ are reals:
$$\begin{align} \phi(f(x))&=\phi(a_0+a_1x+a_2x^2+\cdots a_nx^n)\\ & =a_0+a_1\phi(x)+a_2\phi(x^2)+\cdots a_n\phi(x^n)\\ &=a_0+a_2\phi(x)+a_2\phi(x)^2+\cdots a_n\phi(x)^n=f(\phi(x))\end{align}$$
(I hope the $\phi$ notation is not a problem)
About your second question, this is needed to conclude $\lambda_k=\bar{\lambda_j}$ (i.e., if $\lambda$ it's a root, then its conjugate is a root too).
A: 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 $\overline{f(x)} = f(\overline x)$.
