Factorizing $x^5+1$ as a product of linear and quadratic polynomials. I am encountering some trouble with this question:
Factorize $$x^5+1$$ as a product of real linear and quadratic polynomials.
I know that if we subtract 1 from $x^5+1$, we get that $x^5 = -1$, but I am unsure where to go from here.
Can anyone help with this? Thanks.

If you remember the formula for the sum of a geometric progression, you will have
$$\begin{align} & 1 + x + x^2 + \cdots + x^{n-1} = \frac{1-x^n}{1-x}\\
\iff & x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + \cdots + x + 1) 
\end{align}
$$
When $n = 2k+1 $ is odd, substitute $x$ by $-x$ and move the minus sign around, you will get
$$x^{2k+1} + 1 = (x+1)(x^{2k} - x^{2k-1} + x^{2k-2} + \cdots - x + 1)$$
In particular, for $n = 5$, this gives you
$$x^5 + 1 = (x+1)(x^4 - x^3 + x^2 - x + 1)$$
This sort of factorization is something you should know. If not, you know it now.
Let's look at the coefficients of $x^k$ in second factor, they are $1, -1, 1, -1, 1$. 
Did you notice it is palindromic, i.e stay the same if you reverse the order. For this sort of polynomial. If $r \ne 0$ is a root, so does $\frac1r$. When the number of this list of coefficients is odd, you can simplify the expression by grouping them in powers of $x+\frac1x$. 
Let's see what happens to the factor. We have
$$\begin{align} x^4 - x^3 + x^2 - x + 1
&= x^2(x^2 - x + 1 - x^{-1} + x^{-2})\\
&=  x^2((x^2 + 2 + x^{-2}) - (x + x^{-1}) - 1)\\
&=  x^2((x + x^{-1})^2 - (x + x^{-1}) - 1 )\\
&=  x^2\left[\left(x + x^{-1} -\frac12\right)^2 - \frac{5}{4}\right]\\
\end{align}
$$
Another identity you should remember is for positive $a$, 
$$y^2 - a = (y + \sqrt{a})(y - \sqrt{a})$$
Apply this identity to above expression, you get
$$x^4 - x^3 + x^2 - x + 1
= x^2\left(x + x^{-1} -\frac{1+\sqrt{5}}{2}\right)
\left(x + x^{-1} -\frac{1-\sqrt{5}}{2}\right)$$
Move the $x$ factor around to get rid of the $x^{-1}$ and combine with the $x+1$ factor, one obtain
$$x^5 + 1 = (x+1)\left(x^2 - \frac{1+\sqrt{5}}{2} x + 1 \right)
\left(x^2 - \frac{1-\sqrt{5}}{2} x + 1 \right)$$
This is the decomposition of $x^5+1$ into product of linear and quadratic factors you are seeking.
