Factor \[ x^4+x^3+x+3 \] completely in \(\mathbb F_5[x]\).
Solution
Let \[ f(x)\coloneqq x^4+x^3+x+3\in\FF_5[x]. \]
1. The polynomial \(f\) has no root in \(\FF_5\).
Direct evaluation gives \[ \begin{aligned} f(0)&=3, & f(1)&=1, & f(2)&=4, & f(3)&=4, & f(4)&=2 \end{aligned} \] in \(\FF_5\). Hence \(f\) has no linear factor.
2. If \(f\) were reducible, it would factor as\[ f(x)=(x^2+ax+b)(x^2+cx+d) \]with \(a,b,c,d\in\FF_5\).
Since \(f\) has degree four and has no linear factor by step 1, every nontrivial factorization has factor degrees \(2\) and \(2\). Because \(f\) is monic, multiplying the two factors by inverse nonzero constants makes both quadratic factors monic without changing their product.
3. No factorization from step 2 exists.
Comparing coefficients gives \[ a+c=1, \qquad ac+b+d=0, \qquad ad+bc=1, \qquad bd=3. \] The last equation leaves, up to order, only \[ \{b,d\}=\{1,3\} \qquad\text{or}\qquad \{b,d\}=\{2,4\}. \]
If \(\{b,d\}=\{1,3\}\), then \(b+d=4\), so \[ a+c=1, \qquad ac=1. \] Thus \(a\) and \(c\) would be roots of \[ t^2-t+1. \] Its discriminant is \[ 1-4=2, \] but the squares in \(\FF_5\) are \(0\), \(1\), and \(4\). Hence this case is impossible.
If \(\{b,d\}=\{2,4\}\), then \(b+d=1\), so \[ a+c=1, \qquad ac=4. \] Thus \(a\) and \(c\) are roots of \(t^2-t+4\), whose discriminant is \[ 1-16=0 \] in \(\FF_5\). Therefore \(a=c=3\). But then \[ ad+bc=3(b+d)=3, \] contradicting the required coefficient \(ad+bc=1\).
4. Therefore\[ \boxed{f\text{ is irreducible over }\FF_5}. \]
Step 1 excludes a linear factor, and step 3 excludes a factorization into two quadratics. These are all nontrivial degree patterns for a quartic over a field. Hence \(f\) is irreducible, so its complete factorization in \(\FF_5[x]\) is the polynomial itself.
5. Q.E.D.
Step 4 gives the complete factorization.