x³ − 6x² + 11x − 6 has roots 1, 2 and 3, so the product is 6. The constant term is −6. Reading the constant off the page gives the wrong sign, and it only goes wrong when the degree is odd.
The sum behaves the same way at every degree. Only the product changes its relationship to the constant term, and only when the degree is odd.
Three words for the same fact, and exam questions use all three:
and each one implies the other two. That three-way equivalence is doing more work than it looks: it means you can prove a factor by substituting a number, which is the whole of the factor theorem.
The remainder theorem. Divide p(x) by (x − a) and the remainder is p(a). No long division required.
For p(x) = x³ − 6x² + 11x − 6, the remainder on dividing by (x − 4) is p(4) = 64 − 96 + 44 − 6 = 6. One substitution instead of a page of working.
The factor theorem is the special case where the remainder is zero. p(1) = 1 − 6 + 11 − 6 = 0, so (x − 1) is a factor. That is how you start a cubic you cannot factorise by inspection: try the factors of the constant term, here ±1, ±2, ±3, ±6, and one of them will usually work.
Watch the sign in (x − a). The remainder on dividing by (x + 1) is p(−1) = −24, not p(1). Getting that backwards is the commonest slip in the whole sub-topic, and it is silent: −24 looks like a perfectly good remainder.
For a polynomial anxⁿ + … + a1x + a0 with roots r1, …, rn:
sum of the roots = −an−1 / an
product of the roots = (−1)ⁿ a0 / an
The sum is easy and nobody gets it wrong twice: it is the second coefficient with its sign flipped. The product carries a (−1)ⁿ and that is the entire difficulty.
Where it comes from is worth one line, because then it does not need memorising. Write the polynomial as its factors:
an(x − r1)(x − r2)…(x − rn)
Put x = 0 and every bracket becomes −r. So the constant term is an × (−r1)(−r2)…, which is an times the product of the roots times (−1)ⁿ. Each root contributes one minus sign, so an odd number of roots leaves one over.
Our cubic, where the sign bites. x³ − 6x² + 11x − 6, roots 1, 2 and 3.
Press Next polynomial twice and watch what happens. On the quadratic the constant is 6 and the product is 6: reading it off works. On the quartic the constant is 24 and the product is 24: it works again. It fails only on the cubic, sitting between two cases where it is fine, which is exactly why the habit of reading the constant survives long enough to appear in an exam.
The middle coefficients are in there too. For the cubic, the sum of the products of pairs of roots is 1×2 + 1×3 + 2×3 = 11, which is a₁/a₃. The full set of these is Viète's relations, and Analysis asks only for the sum and the product; but noticing that the 11 is not a coincidence makes the other two formulas look less arbitrary.
The machine finds roots, and from the roots you can check both formulas in two lines. What it will not do is the algebra of an unknown coefficient, which is where these questions usually go.
When you may use it. Analysis Paper 1 is non-calculator, and the sum and product formulas are Paper 1 material precisely because they avoid finding the roots. Paper 2 is where you can check: find the roots, add them, multiply them.
The mark people lose. Giving the constant term as the product of the roots on an odd-degree polynomial. It is right half the time, which is the worst possible frequency: often enough to feel like a rule, not often enough to be one. The habit: write (−1)ⁿ before you write anything else, and for a cubic that is −1, so the constant term's sign flips.
Throughout: p(x) = x³ − 6x² + 11x − 6.
1. Find the sum of the roots.
2. Find the product of the roots.
3. Find the remainder when p(x) is divided by (x − 4).
4. Find the remainder when p(x) is divided by (x + 1).
5. Why does the constant term give the product of the roots for a quartic but not for a cubic?
1 markThe formula written down, with the (−1)ⁿ.
1 markIdentifying an and a0 correctly, including their signs.
1 markThe answer.
On a question that gives you an unknown coefficient and the sum or product of the roots, the formula is what turns it into a linear equation in that unknown. That is the usual shape at Higher Level, and the roots themselves are never found.
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