Regarding complex roots of a polynomial I am having difficulty in finding the roots of the following polynomial when it is given that all the roots are complex:-
$$f(x)= x^4+4x^3+8x^2+8x+4$$
How can I factorize the polynomial to get its roots?
 A: Using binomial theorem;
$$\begin{align}f(x)&=x^4+4x^3+8x^2+8x+4\\
&=\left(x^4+4x^3+6x^2+4x+1\right)+2\left(x^2+2x+1\right)+1\\
&=(x+1)^4+2(x+1)^2+1\\
&=\left((x+1)^2+1\right)^2\\
&=\left(x^2+2x+2\right)^2\end{align}$$
A: Remark that in the general case in $\mathbb R$ a polynomial always get a factorization with smaller polynomials of degree at most $2$.
Since here the leading coefficient is $1$ you search something like $(x^2+ax+b)(x^2+cx+d)$. 
You can develop it and identify the coefficients or try to guess them from the $4's$ and $8's$ in the given expanded polynomial. 
For instance $bd=4$ so you can try $(1,4)$ or $(2,2)$ or $(-1,-4)$, and hopefully $a,c$ will come naturally.
With habit, your factorization skills will refine.
A: Hint:
$$
\begin{align}
f(x)= x^4+4x^3+8x^2+8x+4 & = x^2\left(x^2+\frac{4}{x^2} + 4\left(x + \frac{2}{x}\right) +8\right) \\
 & = x^2\left(\left(x+\frac{2}{x}\right)^2 + 4\left(x + \frac{2}{x}\right) +4\right) \\ 
 & = x^2\left(x+\frac{2}{x} + 2\right)^2 = \;\cdots
\end{align}
$$

[ EDIT ]  The above is based on the observation that the ratio between "symmetric" coefficients of the polynomial $\;\color{blue}1\cdot x^4+\color{red}4\cdot x^3+8\cdot x^2+\color{red}8\cdot x+\color{blue}4 \cdot x^0\;$ follows the powers of $\frac{1}{2}$ from center outwards: $\;\cfrac{8}{8} = \cfrac{1}{2^0}\,, \;\color{red}{\cfrac{4}{8}}=\cfrac{1}{2^1}\,, \;\color{blue}{\cfrac{1}{4}}=\cfrac{1}{2^2}\,$.

This indicates that the polynomial is amenable to a reciprocal polynomial with the substitution $x=\sqrt{2}\,y\,$, and indeed $\color{blue}4\,y^4+\color{red}{8 \sqrt{2}}\,y^3 + 16\,y^2+\color{red}{8\sqrt{2}}\,y+\color{blue}4\,$ is reciprocal.
Lastly, a reciprocal polynomial of even degree $2n$ can be factored as $y^n$ times a polynomial of degree $n$ in $y+\cfrac{1}{y}=\cfrac{1}{\sqrt{2}}\left(x+\cfrac{2}{x}\right)\,$.
This explains why the posted answer started by factoring out $x^2$ then focused on the $x+\cfrac{2}{x}$ terms.
A: If you don't spot something easy first off, you can think that the roots will come in complex conjugate pairs, and will lead to a factorisation into two quadratic factors with real coefficients. You will be able to choose the leading coefficient of each factor to be $1$, so you can try writing what you have as the difference between two squares as follows, with confidence that it will lead to a result:
$$f(x)=(x^2+ax+b)^2-(cx+d)^2$$
To eliminate the $x^3$ term you need $a=2$ and then it is easy to find (in this case) $b=2$ which expresses $f(x)$ as a square.
A: Note that 
$$f(x)=x^4+4x^3+8x^2+8x+4=(x^2+2x+2)^2.$$
Now, the problem is reduced to calculate the roots of a two-degree polynomials, whose roots are $-1-i$ and $-1+i$.
A: It's helpful if you can see how the polynomial factors right away, but if not, you still have a decent method.  You know that complex roots come in conjugate pairs as roots of quadratics, so aim to factor into quadratics.  Moreover you know that the corresponding quadratics have real coefficients, since $$(x-a-bi)(x-a+bi) = x^2 -2ax + a^2 + b^2$$ 
So you can brute force the factor like $$(ax^2 + bx + c)(dx^2 + ex + f) = (x^4 + 4x^3 + 8x^2 + 8x + 4)$$
for $a,b,c,d,e,f$ real.
You then equate coefficients of the resulting monomials:
$$\begin{align}ad &= 1\\
ae + bd &= 4\\
af + be + cd &= 8\\
ec + bf &= 8\\
cf &= 4\end{align}$$
There's a degree of freedom in these equations which arises because you can factor out a scalar from the polynomial factorization.  So start by arbitrarily setting $a = 1$, which implies $d = 1$, leaving us
$$\begin{align}b + e &= 4\\
f + be + c &= 8\\
fb + ec &= 8\\
fc &= 4\end{align}$$
Although I'm personally bad at factoring by eye, I usually notice some symmetry at some point along the way of solving these systems and don't have to go all the way to the end.  Here I want to say $b = e = 2$ by symmetry and $f = c = 2$.  I check that and thankfully it pans out.  
A: One check that is perhaps useful is the one for double roots. In this case one computes
$$
\gcd(f, f') = x^{2} + 2 x + 2
$$
which leads to the factorization already exhibited in the binomial answer.
