Factorization of a 4th degree polynomial 
Factorization of  :
$$x^4-x^3+3x^2+3x+5$$

$$x^4-x^3+3x^2+3x+5=(x^4-x^3+2)+3(x^2+x+1)$$
what do i do ? please help me
 A: A possible error in the question, it might explain things. If we change 5 to 54, we get
$$ x^4 - x^3 + 3x^2 + 3x + 54 = ( x^2 -4 x + 9)  ( x^2 + 3  x + 6)   $$
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First you look for rational roots. There are none. Then we need only check, because of GAUSS, 

The second result states that if a non-constant polynomial with
  integer coefficients is irreducible over the integers, then it is also
  irreducible if it is considered as a polynomial over the rationals.

$$  (x^2 + a x + b) ( x^2 + c x + 5 b),  $$
with $$  b = \pm 1  $$
For both the case $b=1$ and $b=-1,$ we easily get correct coefficients for $x^3$ and $x,$ but then the $x^2$ term comes out to something incorrect. So the thing is irreducible. That is really all that is needed, try two cases.
$$ ( x^2 + x + 1)  ( x^2 - 2  x + 5) = x^4 - x^3 + 4x^2 + 3x + 5 $$
Next comes out non-integer, and still wrong anyway. Gauss says we did not need to continue once $a,c$ came up not being integers.
$$( x^2 - \frac{x}{2} - 1)  (x^2 - \frac{x}{2} - 5)= x^4 - x^3 -\frac{23x^2}{4} + 3x + 5$$
Perhaps I should add that, although there are no real roots, there are real numbers $p,q,r,s$ such that
$$  (x^2 + px+q)(x^2 + rx + s) = x^4 - x^3 + 3 x^2 + 3 x + 5 $$
where both quadratic factors are positive for any chosen $x$
A: This is not an efficient way to show irreducibility, but it's kind of fun:
Theorem:  If $P(x)\in\mathbb{Z}[x]$ is a reducible polynomial of degree $n$, then $|P(m)|$ is prime (or $1$) for at most $2n$ different integers $m$.
Proof:  If $P(x)=A_k(x)B_{n-k}(x)$ where $A_k$ is of degree $1\le k\lt n$ (and $B_{n-k}$ is of degree $1\le n-k\lt n$) and $|P(m)|=|A_k(m)||B_{n-k}(m)|$ is prime (or $1$), then one (or both) of the factors is $1$ or $-1$.  But $A_k$ takes each value at most $k$ times, while $B_{n-k}$ takes value $n-k$ times. So $|A_k(m)|=1$ for at most $2k$ values of $m$ and $|B_{n-k}(m)|=1$ for at most $2(n-k)$ values of $m$, for a total of at most $2n$ possible values of $m$.
So to show that $P(x)=x^4-x^3+3x^2+3x+5$ is irreducible, it suffices to find $9$ integers $m$ such that $|P(m)|$ is prime (or $1$).  A little experimentations does the rest. The following numbers are all prime:
$$\begin{align}
P(0)&=5\\
P(1)&=11\\
P(-1)&=7\\
P(2)&=31\\
P(-3)&=131\\
P(4)&=257\\
P(-6)&=1607\\
P(11)&=13711\\
P(-13)&=31231
\end{align}$$
