how to factor $x^4+2x^3+4x^2+3x+2$ I'm trying my hand on these types of expressions.
How to factorize $x^4+2x^3+4x^2+3x+2$ into two  (or more) polynomials with rational coefficients.  please write step by step solution.
 A: I think the best way is by intuition.
$x^4+2x^3+4x^2+3x+2$
$=(x^4+x^3+x^2)+(x^3+x^2+x)+(2x^2+2x+2)$
$=x^2(x^2+x+1)+x(x^2+x+1)+2(x^2+x+1)$
$=(x^2+x+1)(x^2+x+2)$
A: This works every time:
Assume product of linear and cubic: It is easy to show that this is not the solution you want. 
Assume to be product of 2 quadratics.
$(x^2+ax+b)(x^2+cx+d)$. Compare both sides.
$c+a=2$
$d+ac+b=4$
$ad+dc=3$
$bd=2$
How to solve this: Note that $b,d$ can not be 0. We expect them to be integers (you can call this clever guess as teachers don't want to complicate things).
I will (cleverly) assume $c,a=1$ from the first equation. If I am unable to find a solution, I will proceed to $0,2$. As $a,c=1$, $b+d=3$
As from the last equation, $b,d=1,2 \text{ or } 2,1$. This satisfies the third equation and also the second equation. Bazinga! Note that we assume $a=c$ Hence, the order of $b,d$ does not matter. Never ever try to solve these by substitution unless you want to show the teacher your algebraic capabilities.
Hence:
answer: $(x^2+x+1)(x^2+x+2)$. 
Important Note: For heaven's sake, never write this solution in an exam. Either ask someone the answer or by this method find the answer. And then show teacher your awesome intuitive skills. Like this:
Multiply these brackets in your rough work. 
$x^2(x^2+x+2)+x(x^2+x+2)+1(x^2+x+2)$
$=x^4+x^3+2x^2+x^3+x^2+2x+x^2+x+2$
Now in your fair work, rearrange the question to this, take common factors, and write the answer: 
A: First check for linear factors. In this case (integer coefficients, leading coefficient $=1$) we need only check if $f(x)=0$ for the divisors $\pm1, \pm2$ of the constant term. There is none. Hence we need to check for factors of degree $\ge 2$, which implies that (if at all) the polynomial is the product of two quadratics:
$$ x^4+2x^3+4x^2+3x+2=(x^2+ax+b)(x^2+cx+d)$$
(where wlog. the leading coefficients of the factors are positive and hence $+1$).
Comparing coefficients gives the equations $a+c=2$, $ac+b+d=4$, $ad+bc=3$, $bd=2$. This is not straighforward to solve in general as the equations are nonlinear. But here we can again rest assured that the coefficients are integers, hence it suffices to make a few tests (start from $bd=2$, hence $b,d$ are in $\{-2,-1,1,2\}$)
A: Let $P(x)=x^4+2x^3+4x^2+3x+2$.  If $P$ factors over the rationals, it must factor over the integers.  Since $P$ is monic (i.e., its lead coefficient is $1$), the only possible roots (which would correspond to linear factors, $x-r$) are $\pm1$ and $\pm2$.  We can dismiss $1$ and $2$ out of hand, since $P(x)$ is clearly positive for $x\ge0$.  But $P(-1)=1-2+4-3+2=2$ and $P(-2)=16-16+16-6+2=12$ rules those out as well.  So it remains to see if $P$ can be factored into a pair of quadratics, $P(x)=(x^2+ax+b)(x^2+cx+d)$.
If it can, then $b$ and $d$ are necessarily positive, since otherwise $P$ would have a positive real root, which we know is not the case since, as we already noted, $P(x)$ is positive for $x\ge0$.  (E.g., if $b$ were negative, then the factor $x^2+ax+b$ would be negative at $x=0$ but eventually positive, once $x$ is sufficiently large.)  Thus, from $bd=P(0)=2$, we can restrict ourselves to looking at factorizations of the form
$$P(x)=(x^2+ax+2)(x^2+cx+1)$$
Expanding this out as
$$P(x)=x^4+(a+c)x^3+(ac+3)x^2+(a+2c)x+2$$
we see that we need to have $a+c=2$, $ac+3=4$, and $a+2c=3$.  In general, three equations in two unknowns can not be solved (which is to say, most quartics do not factor into two quadratics), but in this case the solution is pretty clearly $a=c=1$, and that gives the factorization
$$x^4+2x^3+4x^2+3x+2=(x^2+x+2)(x^2+x+1)$$
