Is it possible to express $x^4-x^3+3x^2-4x+6$ as a product of polynomials of smaller degree with integer coefficients? Is it possible to express $x^4-x^3+3x^2-4x+6$ as a product of polynomials of smaller degree with integer coefficients?
My attempt: By equating the polynomial to $0$, one obtains $x=1\pm i, \pm\frac{1}{2}i(\sqrt{11}\mp i)$. From this, one can write the polynomial given in the question as the 
$$(x^2-2x+2)(x^2+x+3)$$
by using Viete formula. 
However, I'm not sure whether I can solve the polynomial before hand and use the answers to express it as product of polynomial of smaller degree. 
 A: Your polynomial has no integer root since such a root must be a divisor of $6$. More over it can have no negative root since, if $x>0$,  $p(x)> 0$. Thus one has to test the values $\,1,2 ,3,6$ and none is a root.
Thus  a factorisation must have the form:
$$x^4-x^3+3x^2-4x+6=(x^2+ax+b)(x^2+a'x+b')$$
which leads to the system:
$$\begin{cases}
a+a'=-1\\aa'+b+b'=3\\ab'+ba'=-4\\bb'=6
\end{cases}$$
Furthermore, $b$ and $b'$ have to be positive: if they were negative, the factors, hence $p(x)$, would have a negative root.
Theere remains two possibilities: $b=2,\enspace b'=3\,$ or $\, b=1,\enspace b'=6$.  From the second equation, we deduce that $aa'=-2$ or $aa'=-4$.
Now we have to solve for the classical problem of finding two numbers, knowing their sum, $-1$ and their product, $-2$ or $-4$. They're roots of one of these equations:
$$t^2+t-2=0,\quad t^2+t-4=0.$$
It happens that the first one has integer roots, $1$ and $-2$, while the second has not.
Thsi third equation will let us know which root is $a$ and which is $a'$. It can be written as $\,3a+2a'=-4$, which enforces $a=-2,\ a'=1$. Finally we obtain the factorisation:
$$x^4-x^3+3x^2-4x+6=(x^2-2x+2)(x^2+ x+3). $$
