Q.Examine for the rational roots of

Step 1. Part (i): $2x^3-x^2-1=0$. $a_n=2,\ a_0=-1$: candidates are $p/q$ with $p\mid1$, $q\mid2$, i.e. $\pm1,\pm\tfrac12$.

Step 2. Test each candidate. $x=1$: $2-1-1=0$ ✓ — a rational root. (The remaining two roots, from dividing out $(x-1)$ to get $2x^2+x+1=0$ with $\Delta=1-8=-7<0$, are non-real.)

Step 3. Part (ii): $x^8-3x+1=0$. Monic ($a_8=1$), $a_0=1$: only candidates are $p\mid1$, i.e. $x=\pm1$.

Step 4. Test each candidate. $x=1: 1-3+1=-1\ne0$. $x=-1: 1+3+1=5\ne0$. Neither works.

Step 5. Conclusion. $2x^3-x^2-1=0$ has the rational root $x=1$; $x^8-3x+1=0$ has no rational roots at all.

Rational Root Theorem — list the finite candidate set from the divisors of $a_0$ and $a_n$, then test each by substitution.

• Forgetting the theorem only narrows candidates — it doesn't guarantee any of them actually work (part ii)
• Missing a candidate by only checking positive divisors