Factoring $x^4+x^2+2x+6$ We have to factor $x^4+x^2+2x+6$.Factoring through factor theorem is not helpful in this question as the question does not follow the integral root theorem i.e. the root of this expression is not any of the factors of 6. I am totally confused.Please help in factoring it!
 A: The polynomial $x^4+x^2+2x+6$ is $x^4+(x+1)^2+5\geqslant5$ for every $x$ hence it has no real root. Thus, there is no factor of degree $1$ hence one can only have two factors of degree $2$, that is, the factorization to be found is $$x^4+x^2+2x+6=(x^2+ax+b)(x^2+cx+d),$$ for some $(a,b,c,d)$. The coefficients of $x^3$ and $x^0$ in the product on the RHS should be $0$ and $6$ respectively hence $a+c=0$ and $bd=6$, which brings us to $$x^4+x^2+2x+6=(x^2+ax+b)(x^2-ax+6/b).$$ Since the coefficients of $x$ and $x^2$ in this product should be $2$ and $1$ respectively, this yields two equations in $(a,b)$. 
Can you write these down and see what happens?
A: $$x^4+x^2+2x+6=x^4+2ax^2+a^2-\left( (2a-1)x^2-2x+a^2-6 \right)$$
Now, $\left( (2a-1)x^2-2x+a^2-6 \right)$  is a constant times perfect square if and only if $\Delta =0$ if and only if
$$4-4(2a-1)(a^2-6)=0$$
if and only if
$$(2a-1)(a^2-6)=1$$
Factoring and using the rational root test you will find that $a=\frac{5}{2}$ is (the only) rational root. 
With this choice of $a$, your polynomial becomes a difference of squares.
A: $\begin{eqnarray}{\bf Hint}\quad x^4\!+x^2\!+2x+6\, &=& \qquad x^4 + 4 &+& x^2\!+2x+2\\
&=&\!\! (x^2\!+\!2)^2\!\!-(2x)^2 &+& x^2\!+2x+2\\
&=&\!\! (x^2\!+\!2 -2x)(\color{#c00}{x^2\!+2 + 2x})\!\! &+& \color{#c00}{x^2\!+2x+2}\\
&=&\!\! (x^2\!-2x+3)(\color{#c00}{x^2\!+ 2x\! + 2})\!\!\!\
\end{eqnarray}$
A: If you are working in $ \mathbb{R}$, note that $p(x)=x^2+2x+6$ and $q(x)=x^4$ are always positive if $x\neq 0$, so its sum is also positive. This means the polynomial has no linear factors.
