Explain why a third-degree polynomial must have exactly one or three real roots. Consider all possibilities and combinations for the $x$-intercepts.

Step 1 ：A third-degree polynomial is of the form \(ax^3 + bx^2 + cx + d = 0\).

Step 2 ：The Fundamental Theorem of Algebra states that every non-zero, single-variable, degree n polynomial with complex coefficients has, counted with multiplicity, exactly n complex roots. This means that a third-degree polynomial must have exactly three roots. However, these roots can be either real or complex.

Step 3 ：A complex root always comes in pairs (conjugate pairs), because if a complex number is a root then its conjugate is also a root. Therefore, a third-degree polynomial can't have two complex roots, because complex roots must come in pairs.

Step 4 ：So, the possibilities for the roots of a third-degree polynomial are: Three real roots or One real root and a pair of complex conjugate roots.

Step 5 ：\(\boxed{\text{Final Answer: A third-degree polynomial must have exactly one or three real roots because of the Fundamental Theorem of Algebra and the fact that complex roots come in conjugate pairs. Therefore, the possible combinations for the roots of a third-degree polynomial are three real roots or one real root and a pair of complex conjugate roots.}}\)