Polynomial Roots — Quadratic, Cubic and Quartic Explained
A polynomial equation asks a simple question: for which values of x does the expression become zero? Those values are the roots. A quadratic has a formula everyone memorises. A cubic has one that almost nobody uses. A quartic has one that is worse still. So what do students actually do in an exam? They use structure — the discriminant, the rational root theorem, factorisation and substitution. This guide shows the practical route for degrees 2, 3 and 4, with every calculation written out.
The rule that governs everything
A polynomial of degree n has exactly n roots, provided you count complex roots and count repeated roots as many times as they repeat. That is the fundamental theorem of algebra. So a cubic always has three roots — you may just not be able to see them all on a real-number graph.
A second rule that saves marks: if the coefficients are all real, then complex roots always come in conjugate pairs. If 2 + 3i is a root, 2 − 3i is one too. That is why a cubic with real coefficients must have at least one real root — three roots cannot pair up evenly.
Degree 2 — the quadratic
x = [ −b ± √(b² − 4ac) ] / 2a with discriminant D = b² − 4ac
| Discriminant | Nature of the roots | Graph of y = ax² + bx + c |
|---|---|---|
| D > 0, a perfect square | Two distinct rational roots | Cuts the x-axis twice |
| D > 0, not a perfect square | Two distinct irrational roots | Cuts the x-axis twice |
| D = 0 | One repeated real root | Touches the x-axis once |
| D < 0 | Two complex conjugate roots | Never touches the x-axis |
Example 1 — 2x² − 7x + 3 = 0
a = 2, b = −7, c = 3.
D = (−7)² − 4(2)(3) = 49 − 24 = 25, and √25 = 5.
x = (7 ± 5) / 4 → x = 12/4 = 3 or x = 2/4 =
0.5
Check by substitution: 2(9) − 7(3) + 3 = 18 − 21 + 3 = 0 ✓ and 2(0.25) − 7(0.5) + 3 = 0.5 − 3.5 + 3 = 0 ✓
Example 2 — complex roots: x² + 2x + 5 = 0
D = 4 − 20 = −16, so √D = 4i.
x = (−2 ± 4i) / 2 = −1 + 2i or
−1 − 2i — a conjugate pair, as expected.
Check: (−1 + 2i)² = 1 − 4i + 4i² = 1 − 4i − 4 = −3 − 4i. Then (−3 − 4i) + 2(−1 + 2i) + 5 = −3 − 4i − 2 + 4i + 5 = 0 ✓
Degree 3 — the cubic
Cardano's formula for a general cubic exists, but in school and entrance exams the intended method is always the same three steps:
- Hunt one root using the rational root theorem. Any rational root p/q of ax³ + bx² + cx + d = 0 has p dividing d and q dividing a. For a monic cubic (a = 1), just test the factors of the constant term: ±1, ±2, ±3 …
- Divide it out. If x = r is a root, then (x − r) is a factor. Divide to get a quadratic.
- Solve the quadratic with the usual formula.
Example 3 — x³ − 6x² + 11x − 6 = 0
Constant term is −6, so test ±1, ±2, ±3, ±6.
At x = 1: 1 − 6 + 11 − 6 = 0 ✓ — so (x − 1) is a factor.
Dividing gives x³ − 6x² + 11x − 6 =
(x − 1)(x² − 5x + 6).
Now x² − 5x + 6 = (x − 2)(x − 3).
Roots: 1, 2 and 3.
Check with Vieta's relations (below): sum = 1 + 2 + 3 = 6 = −b/a ✓; pairwise sum = (1)(2) + (1)(3) + (2)(3) = 2 + 3 + 6 = 11 = c/a ✓; product = 1 × 2 × 3 = 6 = −d/a ✓. All three agree, so the factorisation is right.
Example 4 — a repeated root: x³ − 3x − 2 = 0
At x = 2: 8 − 6 − 2 = 0 ✓
Dividing: x³ − 3x − 2 = (x − 2)(x² + 2x + 1)
= (x − 2)(x + 1)².
Verify the expansion: (x − 2)(x² + 2x + 1) = x³ + 2x² + x − 2x² − 4x − 2 = x³ − 3x − 2 ✓
Roots: 2, −1 and −1. Three roots, but only two distinct values — the graph touches the axis at −1 instead of crossing it.
Vieta's relations — a free check on every answer
Cubic ax³ + bx² + cx + d: sum = −b/a, pairwise sum = c/a, product = −d/a
Quartic ax⁴ + bx³ + …: sum = −b/a, product = e/a (constant term e)
These take ten seconds and catch nearly every sign error. Many entrance questions ask only for the sum or product of the roots — in those you never need to find the roots at all.
The cubic discriminant
For ax³ + bx² + cx + d, the discriminant is
Δ > 0 means three distinct real roots; Δ = 0 means a repeated root; Δ < 0 means one real root and two complex conjugates. Test it on Example 3 (a = 1, b = −6, c = 11, d = −6): 18abcd = 18 × 396 = 7128; −4b³d = −4 × (−216)(−6) = −5184; b²c² = 36 × 121 = 4356; −4ac³ = −4 × 1331 = −5324; −27a²d² = −27 × 36 = −972. Adding: 7128 − 5184 + 4356 − 5324 − 972 = 4, which is positive — three distinct real roots, exactly what we found. On Example 4 (a = 1, b = 0, c = −3, d = −2) the same formula gives 0 + 0 + 0 + 108 − 108 = 0, correctly flagging the repeated root.
Degree 4 — the quartic
A general quartic can be solved by Ferrari's method, but exams almost never want that. What they do want is the biquadratic substitution: if the equation contains only even powers, put y = x².
Example 5 — x⁴ − 5x² + 4 = 0
Let y = x². Then y² − 5y + 4 = 0 → (y − 1)(y − 4) = 0 → y = 1 or y = 4.
Back-substitute: x² = 1 → x = ±1; x² = 4 →
x = ±2.
Roots: 1, −1, 2, −2 — four roots for a degree-4 equation, as required.
Check x = 2: 16 − 5(4) + 4 = 16 − 20 + 4 = 0 ✓ Check x = −1: 1 − 5 + 4 = 0 ✓
The other standard quartic route is factorising into two quadratics, often after spotting a rational root with the same theorem used for cubics. And a fact worth knowing: for degree 5 and above no general formula in radicals exists at all — that is the Abel–Ruffini theorem. Numerical methods are not a shortcut there; they are the only option.
Common mistakes
- Losing roots in the biquadratic substitution. After finding y you must take both square roots. Stopping at y = 1, 4 and writing "roots are 1 and 4" is the single most common error in this topic.
- Sign slip in the quadratic formula. It is −b, not b. With b = −7 the numerator starts at +7.
- Saying "no roots" when D < 0. There are still two roots; they are complex. Only say "no real roots".
- Testing only positive factors in the rational root theorem. Negative candidates matter — in x³ − 3x − 2 the repeated root is −1.
- Forgetting multiplicity. (x + 1)² contributes two roots. A question asking "how many roots" wants three for a cubic, counted with multiplicity.
- Cancelling a factor containing x. Dividing x³ = 4x by x silently throws away the root x = 0. Factorise instead.
Where this appears in exams
| Exam / class | Typical use |
|---|---|
| CBSE/ICSE Class 10–12 | Quadratic equations, nature of roots, polynomial factorisation |
| JEE Main / Advanced | Conditions on the discriminant, sum and product of roots, common-root problems |
| Class 11–12 chemistry | Quadratics from Ka, Kb and Kc problems; cubics from solubility products |
| IIT-JAM / GATE / CSIR-NET | Characteristic (secular) equations in quantum chemistry and matrix problems |
For the exact syllabus and marking scheme, always read the current official notification of the exam you are writing.
Confirm your factorisation. Enter the coefficients and the polynomial root finder returns every root — real and complex — so you can check the roots you obtained by hand before writing them in an answer sheet.
Open the Polynomial Roots Calculator →Quadratics from equilibrium and solubility problems trip up more Class 11–12 students than the chemistry itself. ABC Chemistry teaches Class 11–12 chemistry at the Gurugram centre and online across India — abcchemistry.in.