Factor x^3 + x^2 + 2x + 1, or show that it does not factor over the rationals.

Step-by-step solution

1. 1

   Reduce the candidate list. The leading coefficient is $1$ and the constant is $1$, so any rational root $p/q$ has $p\mid 1$ and $q\mid 1$. Only two candidates survive:

   $x=1\quad\text{and}\quad x=-1$

2. 2

   Test both. With $f(x)=x^3+x^2+2x+1$:

   $f(1)=1+1+2+1=5\neq 0$
   $f(-1)=-1+1-2+1=-1\neq 0$

   Neither is a root, so $f$ has no linear factor with rational coefficients.

3. 3

   Explain why that settles it for a cubic. Any factorisation of a degree-3 polynomial over a field must include a degree-1 factor (the only way to split $3$ is $1+2$ or $1+1+1$). Having ruled out all linear rational factors, $f$ is irreducible over $\mathbb{Q}$.

4. 4

   Show why grouping also fails. The natural pairing gives

   $x^2(x+1)+1(2x+1)$

   and the brackets $(x+1)$ and $(2x+1)$ are different, so no common binomial can be extracted. Grouping succeeds only when the two brackets coincide, as they do in $x^3+x^2-x-1$.

5. 5

   Locate the single real root numerically for completeness. $f(-0.5)=0.125$ and $f(-0.6)=-0.056$, so the real root lies near $x\approx-0.5698$; the other two roots are a complex conjugate pair. Being irrational, that root cannot give a rational factor $\checkmark$.