To determine which situation cannot be solved using the quadratic formula, we need to recall that the quadratic formula is used for solving equations of the form $ax^2 + bx + c = 0$. The formula is:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Let's analyze each option:

• A. One term of the polynomial has a degree of 3.

  The quadratic formula only applies to equations where the highest degree of the variable is 2. If a polynomial has a term with a degree of 3, it is a cubic equation, not a quadratic equation, and thus cannot be solved using the quadratic formula.

• B. The right-hand side of the equation is zero.

  Having the right-hand side of the equation be zero is actually the standard form of a quadratic equation ($ax^2 + bx + c = 0$), so this does not prevent the use of the quadratic formula.

• C. The coefficient of the $x^2$-term is 1.

  If the coefficient of the $x^2$-term is 1, the quadratic equation can still be solved using the quadratic formula. In fact, this simplifies the equation but does not prevent the use of the formula.

• D. The coefficient of the $x$-term is zero.

  If the coefficient of the $x$-term is zero, the equation becomes $ax^2 + c = 0$. This is still a quadratic equation and can be solved using the quadratic formula, where $b = 0$.

The situation where the quadratic formula cannot be used is:

$\boxed{\text{A. One term of the polynomial has a degree of 3.}}$

Do you want details or have any questions? Here are five more questions you might ask:

1. How do you solve a cubic equation if the quadratic formula cannot be used?
2. What are the steps to derive the quadratic formula?
3. Can the quadratic formula be used for complex solutions?
4. What is the significance of the discriminant in the quadratic formula?
5. How do you solve a quadratic equation by factoring?

Tip: Remember that the quadratic formula is specifically designed for equations of the form $ax^2 + bx + c = 0$. For higher-degree polynomials, different methods or formulas are required.