Factoring polynomials in Q[x] Factor $x^4+x^2+1$ in $Q[x]$. I am slightly confused how to go about doing this. Is there someway besides using basic algebra and completing the square to do this problem in the context of modern algebra? 
 A: Generally when factoring a polynomial one tries to reduce the number of potential factors that need to be considered as much as possible. There are several methods and tricks used to eliminate possibilities, some of which were mentioned in the comments. A brute force solution is avoided as much as possible, and used only after as many types of factors as possible have been eliminated.   
In your specific case we can reason as follows: by the rational root test we see that only $\pm 1$ are possible roots. After checking that neither is actually a root we know that if $ x^4+x^2+1$ is reducible, then its factors must be polynomials of second degree, i.e. $$x^4+x^2+1=(x^2+ax+b)(x^2+cx+d).$$  
By the Gauss lemma we need only test integers values for $a,b,c,d$. Now we can see that $bd=1$ so we have that either $b=d=1$ or $b=d=-1$. Since the coefficient of $x^3$ must be zero in the product, we have that $a+c=0$. Since the coefficient of $x$ must also be $0$, we have that $ad+bc=0$. These restrictions limit our options for factors that need to be considered. Now an obvious first pick would be $a=1, c=-1$ and $b=d=1$, which satisfy these restrictions and so they're a candidate solution. Testing them we see that yes, the equation is satisfied 
$$x^4+x^2+1=(x^2+x+1)(x^2-x+1).$$    
