Reading the constant term as the product of the roots works for every even degree and fails for every odd one. That is often enough to feel like a rule and not often enough to be one, which is why it survives until an exam.
Ask for the product of the roots on each of the three views before revealing anything. On the quadratic the class will say 6 and be right. On the cubic most will say −6 and be wrong. On the quartic they will say 24 and be right again.
That sequence is the whole lesson, and it is worth getting the wrong answer on the board in between two right ones. A rule that fails in the middle of a run of successes is almost impossible to unlearn by being told; it has to be caught.
The red dot on the cubic view sits at (0, −6), where the curve crosses the y-axis, and the three green dots sit at 1, 2 and 3. Point at all four at once: the constant term is a point on the graph and the product of the roots is a property of three other points. There is no reason they should be equal, and the surprise is that they ever are.
| Question | Answer |
|---|---|
| 1. Sum of the roots | −(−6)/1 = 6. |
| 2. Product of the roots | (−1)³(−6)/1 = 6. |
| 3. Remainder by (x − 4) | p(4) = 6. |
| 4. Remainder by (x + 1) | p(−1) = −24. |
| 5. Why even works and odd does not | B. Each root contributes a minus sign at x = 0. |
Questions 1 and 2 both have the answer 6, which is deliberate: the sum and the product of 1, 2 and 3 happen to coincide. It stops a student checking their method by whether the two answers look different, and it makes the −6 answer on question 2 stand out as a decision rather than a slip.
1 markThe formula written down, with the (−1)ⁿ.
1 markIdentifying an and a0 with their signs.
1 markThe answer.
The usual Higher Level shape is not "find the product" but "the product of the roots is 12, find k". Then the formula turns the question into a one-line equation, and the mark is for setting it up, not for solving it. Teach it in that direction as well as forwards, because that is the direction the paper asks.
| They wrote | What happened |
|---|---|
| −6 for the product | The error this page exists for. Read the constant term off. Do not explain it; press Next polynomial twice and let the quartic undermine their rule. |
| −6 for the sum | Forgot the minus in −a₂/a₃. Different error, same number. Ask which formula they used. |
| 11 for either | Used a₁ instead of a₂. For the product it is also the sum of the products of pairs, which is a right answer to a question nobody asked. |
| 1 | Gave a₃, the leading coefficient. Usually a reading error in the formula rather than a misconception. |
| 0 on question 3 | Assumed (x − 4) is a factor. Worth asking why they thought so: often they have spotted that 4 divides into the constant 6's neighbours and over-applied the rational root idea. |
| −24 on question 3 | Substituted −4. The sign confusion, one question early. |
| +24 on question 4 | Right size, wrong sign. p(−1) = −1 − 6 − 11 − 6, and every single term is negative, so a positive answer means they have not written the substitution out. |
| p(1) for (x + 1) | The quiet error. It gives 0, so they conclude (x + 1) is a factor, which is false. Insist on writing (x + 1) = (x − (−1)) in full the first few times. |
"Why is there a (−1)ⁿ but not on the sum?" Because the sum comes from adding the roots and the product comes from multiplying them, and multiplying is where the minus signs accumulate. Do the x = 0 substitution on the board for a quadratic and a cubic side by side: (−r₁)(−r₂) has two minuses and (−r₁)(−r₂)(−r₃) has three. That is the entire proof and it takes a minute.
"What if the leading coefficient is not 1?" Then both formulas divide by it, which is what the an in the denominator is for. Give them 2x³ − 12x² + 22x − 12, the same cubic doubled: the roots are unchanged at 1, 2 and 3, the constant is now −12, and (−1)³(−12)/2 is still 6. Students who have only met monic polynomials quietly drop the division.
"Do these work with complex roots?" Yes, and that is the reason they are useful. x³ − 1 has one real root and two complex ones; the sum is 0, which is −0/1, and the product is 1, which is (−1)³(−1)/1. The formulas never ask whether the roots are real, which is why they can answer questions where finding the roots would be impossible by hand.
"How do I find the first factor of a cubic?" Try the factors of the constant term divided by the factors of the leading one. For this cubic that is ±1, ±2, ±3, ±6 and one of them works immediately. Say that it is a search and not a formula, and that an exam will always make the first one findable.
| Demonstrate | Define p once on the Nspire, then put p(4), p(1) and p(-1) on three consecutive lines: 6, 0, -24. Three remainders in three keystrokes, against a page of long division each. That contrast is what sells the remainder theorem. |
|---|---|
| Where they stick | The superscript box. Typing x^3-6*x^2+11*x-6 puts everything after the 3 inside the exponent and defines something else entirely, with no error message. Use the right arrow after the 3, or the squared key for the 2. |
| The check | On the Casio, the Equation app's Polynomial solver takes the coefficients straight from the question and returns 1, 2, 3. Then 1+2+3 and 1*2*3 confirm both formulas numerically. |
The Casio route is genuinely faster here than the Nspire's, because the question hands you coefficients and the Polynomial solver wants coefficients. Worth telling a mixed-machine class which of them has the shortcut on this sub-topic.
| Step | What |
|---|---|
| 1 | Zero, root and factor as three names for one fact. Get them to state all three about x = 1. |
| 2 | The remainder theorem by doing one long division and then one substitution. The comparison does the arguing. |
| 3 | The factor theorem as the remainder being zero. Find the first factor of the cubic by search. |
| 4 | Factorise fully, collect the roots 1, 2 and 3. |
| 5 | Ask for the sum and the product from the roots. Both 6. |
| 6 | Now the formulas. Derive them from the factorised form at x = 0. |
| 7 | The figure, all three views, product first and explanation second. |
| 8 | Backwards: "the product of the roots is 12, find k". That is the exam shape. |
Do not say "the product of the roots is the constant term". It is said in nearly every classroom because it is true for the quadratics students met first, and it is the direct cause of the −6 answer two years later. If you want a short version: the constant term, with the sign flipped for an odd degree.
Do not skip the non-monic case. Every textbook example has a leading coefficient of 1, so the an in the denominator looks like decoration until a paper puts a 2 there. One worked example with 2x³ takes three minutes and prevents a whole category of lost marks.