Chapter 3.4 — Exercise 3.3 — Zeroes and Coefficients
Relation between zeroes and coefficients of a polynomial. This is Lesson 4 of 5 in Chapter 3: Polynomials.
A Shortcut Around Solving
Finding a polynomial's zeroes usually means factorising and solving. Exercise 3.3 shows there's a faster route in the other direction too: once you know a polynomial's zeroes, its coefficients follow from two or three simple formulas — no factorisation required at all, and the same formulas run just as well backward, from coefficients straight to zero-sums and zero-products.
Quadratic Zeroes, Sum and Product
Take p(x) = 3x² + 5x − 2, which factorises to (x + 2)(3x − 1), giving zeroes −2 and 1/3. Their sum is −2 + 1/3 = −5/3, and their product is −2 × 1/3 = −2/3 — both of which match a direct formula involving nothing more than the coefficients themselves:
Sum of zeroes: α + β = −b/a Product of zeroes: αβ = c/aFor 3x² + 5x − 2 (a = 3, b = 5, c = −2): −b/a = −5/3 and c/a = −2/3 — exactly matching the sum and product computed directly from the factored zeroes. This is genuinely useful in reverse, too: given only a, b, and c, you can state the sum and product of a quadratic's zeroes without ever factorising it, solving for it, or knowing what the individual zeroes actually turn out to be.
Extending to Cubics
The same idea extends to a cubic ax³ + bx² + cx + d with three zeroes α, β, γ, except now there are three relationships instead of two:
α + β + γ = −b/a αβ + βγ + γα = c/a αβγ = −d/aFor p(x) = 2x³ − 5x² − 14x + 8, which factorises to (2x − 1)(x + 2)(x − 4), the zeroes are 1/2, −2, and 4. Their sum is 5/2, matching −b/a = −(−5)/2; the sum of products taken two at a time is −7, matching c/a = −14/2; and their product is −4, matching −d/a = −8/2. The middle relationship — αβ + βγ + γα — is the one most often mishandled, since it's easy to forget one of the three cross-terms partway through, especially once negative zeroes are involved and the signs start stacking up.
Four Cubics, Read Straight From Coefficients
Once the three formulas are known, all three quantities can be found for any cubic without factorising it at all — just by reading off a, b, c, and d:
| Polynomial | α+β+γ | αβ+βγ+γα | αβγ |
|---|---|---|---|
| x³ + 3x² − x − 2 | −3 | −1 | 2 |
| 4x³ + 8x² − 6x − 2 | −2 | −3/2 | 1/2 |
| x³ + 4x² − 5x − 2 | −4 | −5 | 2 |
| x³ + 5x² + 0x + 4 | −5 | 0 | −4 |
The last row is the one to watch: x³ + 5x² + 4 has no x-term at all, which means c = 0 — and the "sum of products two at a time" comes out to exactly 0/1 = 0 as a direct result, not a coincidence.
Six Quadratics, Zeroes Found and Verified
A recurring question type asks you to factorise a quadratic to find its zeroes, then confirm the sum/product formulas hold for each one. Splitting the middle term is the standard route to the two factors:
| Polynomial | Zeroes | Sum (matches −b/a) | Product (matches c/a) |
|---|---|---|---|
| x² − 2x − 8 | 4, −2 | 2 | −8 |
| 4s² − 4s + 1 | 1/2, 1/2 | 1 | 1/4 |
| 6x² − 7x − 3 | 3/2, −1/3 | 7/6 | −1/2 |
| 4u² + 8u | 0, −2 | −2 | 0 |
| t² − 15 | √15, −√15 | 0 | −15 |
| 3x² − x − 4 | 4/3, −1 | 1/3 | −4/3 |
4s² − 4s + 1 is worth a closer look: splitting the middle term gives 4s² − 2s − 2s + 1 = 2s(2s − 1) − (2s − 1) = (2s − 1)², so both zeroes coincide at s = 1/2 — the quadratic equivalent of the "touches but doesn't cross" parabola from the previous exercise. t² − 15 is worth a second look too: with no middle term at all, splitting isn't needed — the expression is already a difference of squares (t² − (√15)²), so it factorises directly to (t − √15)(t + √15) without any splitting step.
Building a Polynomial From Scratch
The reverse process — given only the sum and product of the zeroes, construct the polynomial that produces them — uses one general formula:
Required polynomial = k[x² − (sum of zeroes)x + (product of zeroes)], for any non-zero kMultiplying by a constant k never changes a polynomial's zeroes, so k is typically chosen just to clear away fractions and leave whole-number coefficients behind — a matter of convenience and readability, not correctness, since technically any non-zero value of k produces an equally valid answer.
| Sum, Product given | k chosen | Resulting polynomial |
|---|---|---|
| 1/4, −1 | 4 | 4x² − x − 4 |
| √2, 1/3 | 3 | 3x² − 3√2x + 1 |
| 0, √5 | 1 | x² + √5 |
| 1, 1 | 1 | x² − x + 1 |
| −1/4, 1/4 | 4 | 4x² + x + 1 |
| 4, 1 | 1 | x² − 4x + 1 |
A related question runs the same formula with the zeroes given directly instead of their sum and product: for zeroes 2 and −1, the polynomial is x² − x − 2; for √3 and −√3, it's x² − 3 (the sum vanishes, leaving no middle term at all); for 1/4 and −1, choosing k = 4 gives 4x² + 3x − 1; and for 1/2 and 3/2, choosing k = 4 gives 4x² − 8x + 3. Notice that √3 and −√3 producing a middle-term-free polynomial isn't a special property of square roots — any pair of zeroes that are negatives of each other will always cancel out to a zero sum, whole numbers included.
Verifying All Three Relationships on a Cubic
The closing question asks you to confirm directly that 1, −1, and 3 are zeroes of x³ − 3x² − x + 3, then check all three cubic formulas at once. Substituting each value directly: p(1) = 1 − 3 − 1 + 3 = 0, p(−1) = −1 − 3 + 1 + 3 = 0, and p(3) = 27 − 27 − 3 + 3 = 0 — all three check out.
Comparing the polynomial to ax³ + bx² + cx + d gives a = 1, b = −3, c = −1, d = 3. The sum of the zeroes, 1 + (−1) + 3 = 3, matches −b/a = 3; the sum of products taken two at a time, (1)(−1) + (−1)(3) + (3)(1) = −1, matches c/a = −1; and the full product, (1)(−1)(3) = −3, matches −d/a = −3. All three relationships hold simultaneously, exactly as the general formulas predict — a satisfying full-circle check, since this single example verifies every formula introduced earlier in the exercise at once, rather than just one at a time.
Continuing On
The sum/product formulas used throughout this exercise depend entirely on already knowing a polynomial's zeroes — Exercise 3.4 tackles the harder direction, dividing one polynomial by another to find unknown zeroes when only a few are given upfront and the rest must be recovered through division rather than substitution. For the graphical picture behind why these zeroes exist at all, see Exercise 3.2.