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.

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
$$x^4+2x^3+4x^2+3x+2=(x^2+x+2)(x^2+x+1)$$
