Remainder theorem: if a polynomial f(x) leaves remainder r when divided by (x−a), then f(a) = r. Testing each option by evaluating at x=5, x=3, and x=2 checks which one gives remainder 1 in all three cases.
Testing option 2, f(x)=x³−10x²+31x−29: f(5)=125−250+155−29=1 ✓; f(3)=27−90+93−29=1 ✓; f(2)=8−40+62−29=1 ✓. All three conditions are satisfied by this polynomial.
Full option-by-option analysis available to enrolled students.