<s>[INST] 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
 [/INST] 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 $x$
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