In short

A polynomial equation of degree n is a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0, with a_n \neq 0. The Fundamental Theorem of Algebra guarantees it has exactly n roots in the complex numbers, counted with multiplicity. The sum of those roots, the sum of their pairwise products, and so on, are all locked into the coefficients — a pattern that generalises what you already know from quadratics.

You have solved quadratic equations. A quadratic has degree 2, and it always has exactly 2 roots — sometimes both real, sometimes a repeated pair, sometimes a complex conjugate pair. But always two.

Now take a cubic: x^3 - 6x^2 + 11x - 6 = 0. Try x = 1: 1 - 6 + 11 - 6 = 0. Try x = 2: 8 - 24 + 22 - 6 = 0. Try x = 3: 27 - 54 + 33 - 6 = 0. Three roots — 1, 2, 3 — for a degree-3 equation.

Is this a coincidence, or is there a pattern?

The pattern is one of the deepest facts in all of mathematics: a polynomial equation of degree n always has exactly n roots, once you count them the right way. "The right way" means allowing complex numbers and counting repeated roots according to their multiplicity. This fact is called the Fundamental Theorem of Algebra, and this article is about what it says, why you should believe it, and what it lets you do.

What a polynomial equation is

A polynomial equation is an equation of the form

a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0

where a_0, a_1, \ldots, a_n are constants (real or complex), and a_n \neq 0. The number n is the degree of the equation. You have already met the special cases:

Degree Name General form Number of roots
1 Linear ax + b = 0 1
2 Quadratic ax^2 + bx + c = 0 2
3 Cubic ax^3 + bx^2 + cx + d = 0 3
4 Quartic ax^4 + \cdots = 0 4
n General a_n x^n + \cdots + a_0 = 0 n

The last column is the claim. Every row says: degree n, exactly n roots. The first two rows you can verify from formulas you already know. The general claim is the Fundamental Theorem.

Polynomials of degrees 1 through 4 and the number of times each crosses the x-axisFour small graphs arranged side by side. A linear function crosses the axis once. A quadratic crosses twice. A cubic crosses three times. A quartic crosses four times. Each graph has red dots marking the roots. degree 1 1 root degree 2 2 roots degree 3 3 roots degree 4 4 roots
As the degree increases, so does the maximum number of times the curve can cross the horizontal axis. A degree-$n$ polynomial crosses at most $n$ times — and over the complex numbers, it always has exactly $n$ roots (counted with multiplicity).

The Fundamental Theorem of Algebra

The theorem can be stated in one sentence.

Fundamental Theorem of Algebra

Every polynomial equation of degree n \geq 1 with complex coefficients has exactly n roots in the complex numbers, counted with multiplicity.

"Counted with multiplicity" means: if (x - r)^k divides the polynomial but (x - r)^{k+1} does not, then r is counted k times. The equation (x - 2)^3 = 0 has degree 3 and one distinct root (x = 2), but that root has multiplicity 3, so the count is 3.

The theorem says something remarkable: no matter how high the degree, no matter how complicated the coefficients, the polynomial always has enough roots. It cannot "run out" of solutions the way x^2 + 1 = 0 runs out of real solutions. Over the complex numbers, every polynomial of degree n factors completely:

a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0 = a_n(x - r_1)(x - r_2) \cdots (x - r_n)

where r_1, r_2, \ldots, r_n are the n roots (some possibly equal, some possibly complex).

The proof of this theorem is not elementary — it requires tools from complex analysis or topology that go well beyond school algebra. But the statement is something you can use right now, and it shapes every calculation in this article.

A consequence for real polynomials

If all the coefficients a_0, a_1, \ldots, a_n are real numbers, the complex roots come in conjugate pairs. That is, if \alpha + \beta i is a root (with \beta \neq 0), then \alpha - \beta i is also a root. This means a real polynomial of odd degree always has at least one real root — because the complex roots pair off, and an odd number of roots cannot all be paired.

Complex roots of real polynomials come in conjugate pairsA diagram showing a degree-4 polynomial with roots r1 and r2 real, and the pair alpha plus beta i and alpha minus beta i as conjugate complex roots. A bracket groups the two complex roots together. degree 4: roots are r₁ , r₂ , (α + βi) , (α − βi) conjugate pair real coefficients → complex roots always pair up
When all coefficients are real, complex roots are never alone — they come in conjugate pairs. A degree-4 equation might have four real roots, or two real and one conjugate pair, or two conjugate pairs. But never one or three complex roots.

Relation between roots and coefficients

Here is where the theorem becomes a working tool. Suppose the polynomial equation

a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0

has roots r_1, r_2, \ldots, r_n. By the Fundamental Theorem, the polynomial factors as

a_n(x - r_1)(x - r_2) \cdots (x - r_n) = 0

Expand the right side and compare coefficients. What falls out is a remarkable set of identities.

The pattern for any degree

Divide both sides by a_n to get the monic form (leading coefficient 1):

x^n + \frac{a_{n-1}}{a_n} x^{n-1} + \cdots + \frac{a_1}{a_n} x + \frac{a_0}{a_n} = (x - r_1)(x - r_2) \cdots (x - r_n)

Expanding (x - r_1)(x - r_2) \cdots (x - r_n) is a systematic process. The coefficient of x^{n-1} comes from choosing "x" from all but one factor and "-r_i" from the remaining one. The coefficient of x^{n-2} comes from choosing "-r_i" from two factors and "x" from the rest. And so on.

The result, comparing the left side with the expanded right side:

\sum_{i} r_i = r_1 + r_2 + \cdots + r_n = -\frac{a_{n-1}}{a_n}
\sum_{i < j} r_i r_j = r_1 r_2 + r_1 r_3 + \cdots + r_{n-1} r_n = \frac{a_{n-2}}{a_n}
\sum_{i < j < k} r_i r_j r_k = -\frac{a_{n-3}}{a_n}
\vdots
r_1 r_2 \cdots r_n = (-1)^n \frac{a_0}{a_n}

The signs alternate: the sum of roots has a - sign, the sum of pairwise products has a +, the sum of triple products has a -, and so on.

Checking against what you know

For a quadratic ax^2 + bx + c = 0 with roots \alpha and \beta:

These are exactly the formulas you already know from quadratic equations. The general relations are the same idea, extended to higher degrees.

For a cubic ax^3 + bx^2 + cx + d = 0 with roots \alpha, \beta, \gamma:

Root-coefficient relations for quadratic, cubic, and quartic equationsA table showing the relation between roots and coefficients for equations of degree 2, 3, and 4. For degree 2, the sum of roots is negative b over a and the product is c over a. For degree 3, there are three relations. For degree 4, there are four. Degree Sum of roots Product of roots 2 −b/a c/a 3 −b/a −d/a 4 −b/a e/a n −aₙ₋₁/aₙ (−1)ⁿ a₀/aₙ the sum always picks up a minus sign; the product alternates
The sum and product of the roots for degrees $2$, $3$, $4$, and the general degree $n$. The pattern is the same at every level: the sum of roots is $-a_{n-1}/a_n$, and the product of all roots is $(-1)^n a_0/a_n$.

Why the sign pattern alternates

Here is the intuition. Start with two roots and expand:

(x - r_1)(x - r_2) = x^2 - (r_1 + r_2)x + r_1 r_2

The coefficient of x picks up a minus sign (because each -r_i contributes one negative). The constant term is a product of two negatives, so it is positive. Now add a third root:

(x - r_1)(x - r_2)(x - r_3) = x^3 - (r_1 + r_2 + r_3)x^2 + (r_1 r_2 + r_1 r_3 + r_2 r_3)x - r_1 r_2 r_3

Each time you pick one more (-r_i) from the factors, you introduce one more minus sign. Picking one r_i gives (-1)^1, picking two gives (-1)^2 = +1, picking three gives (-1)^3 = -1. The signs alternate because each additional root chosen flips the sign once.

Two worked examples

Example 1: Find the sum and product of roots of $2x^3 - 9x^2 + 12x - 4 = 0$ without solving

Step 1. Identify the coefficients. Here a_3 = 2, a_2 = -9, a_1 = 12, a_0 = -4.

Why: the leading coefficient is a_n = a_3 = 2, and the root-coefficient relations all divide by a_n.

Step 2. Compute the sum of roots.

r_1 + r_2 + r_3 = -\frac{a_2}{a_3} = -\frac{-9}{2} = \frac{9}{2}

Why: the sum of roots formula -a_{n-1}/a_n gives -(-9)/2 = 9/2. The two minus signs cancel.

Step 3. Compute the sum of pairwise products.

r_1 r_2 + r_1 r_3 + r_2 r_3 = \frac{a_1}{a_3} = \frac{12}{2} = 6

Why: for a cubic, the sum of pairwise products is a_{n-2}/a_n = a_1/a_3, with a positive sign.

Step 4. Compute the product of all three roots.

r_1 r_2 r_3 = -\frac{a_0}{a_3} = -\frac{-4}{2} = 2

Why: the product formula for a cubic is (-1)^3 a_0/a_3 = -a_0/a_3 = -(-4)/2 = 2.

Result. Sum = 9/2, pairwise product sum = 6, product = 2.

Interactive graph of 2x cubed minus 9x squared plus 12x minus 4A coordinate plane showing the cubic curve y equals 2x cubed minus 9x squared plus 12x minus 4. The curve crosses the x-axis at three points. A draggable red point on the curve lets you trace the function value as you move along it. x y 1 2 3 1 −1 drag the red point
The cubic $2x^3 - 9x^2 + 12x - 4$. Drag the red point to see where the curve crosses zero — those crossings are the roots. Their sum is $9/2$, the sum of their pairwise products is $6$, and their product is $2$, all readable from the coefficients alone.

You can verify by factoring: 2x^3 - 9x^2 + 12x - 4 = 2(x - 1/2)(x - 2)(x - 2) = (2x - 1)(x - 2)^2. The roots are 1/2, 2, 2. Check: 1/2 + 2 + 2 = 9/2. (1/2)(2) + (1/2)(2) + (2)(2) = 1 + 1 + 4 = 6. (1/2)(2)(2) = 2. Everything matches.

Example 2: A quartic equation has roots $1$, $-1$, $2$, $-3$. Find the equation.

Step 1. Use the factored form. The quartic with these roots and leading coefficient 1 is

p(x) = (x - 1)(x + 1)(x - 2)(x + 3)

Why: by the Fundamental Theorem, a monic degree-4 polynomial with roots r_1, \ldots, r_4 is (x - r_1)(x - r_2)(x - r_3)(x - r_4).

Step 2. Multiply the first pair: (x - 1)(x + 1) = x^2 - 1.

Why: the difference-of-squares identity (a - b)(a + b) = a^2 - b^2 makes this quick.

Step 3. Multiply the second pair: (x - 2)(x + 3) = x^2 + 3x - 2x - 6 = x^2 + x - 6.

Why: standard FOIL expansion.

Step 4. Multiply the two quadratics: (x^2 - 1)(x^2 + x - 6).

= x^4 + x^3 - 6x^2 - x^2 - x + 6
= x^4 + x^3 - 7x^2 - x + 6

Why: every term of the first factor distributes across every term of the second, then collect like terms.

Result. The equation is x^4 + x^3 - 7x^2 - x + 6 = 0.

Graph of the quartic x to the fourth plus x cubed minus 7x squared minus x plus 6 with roots at negative 3, negative 1, 1, and 2A coordinate plane showing a quartic curve that crosses the x-axis at x equals negative 3, negative 1, 1, and 2 — four roots for a degree-4 equation. x y 1 2 −1 −2 −3 −3 −1 1 2
The quartic $x^4 + x^3 - 7x^2 - x + 6$ crosses the axis at exactly four points: $x = -3$, $-1$, $1$, $2$. Four roots for a degree-$4$ equation — the Fundamental Theorem in action. Check the root-coefficient relations: sum of roots $= -3 + (-1) + 1 + 2 = -1 = -a_3/a_4 = -1/1$, and product $= (-3)(-1)(1)(2) = 6 = (-1)^4 a_0/a_4 = 6/1$.

Verify the root-coefficient relations. Sum of roots: 1 + (-1) + 2 + (-3) = -1. The coefficient of x^3 is 1, and -a_3/a_4 = -1/1 = -1. Match. Product of roots: (1)(-1)(2)(-3) = 6. The constant term is 6, and (-1)^4 a_0/a_4 = 6/1 = 6. Match.

Common confusions

Going deeper

If you came here to learn what polynomial equations are, what the Fundamental Theorem says, and how roots relate to coefficients, you have all of it — you can stop here. What follows is for readers who want the broader view.

Why "fundamental"?

The theorem is called "fundamental" not because its proof is basic — the proof is genuinely hard — but because it is the foundation on which the entire theory of polynomials rests. Without it, you could not be sure that factoring a polynomial into linear factors is always possible. Every factoring method you use — the Factor Theorem, synthetic division, grouping — implicitly relies on the Fundamental Theorem guaranteeing that the roots exist in the first place.

The theorem was first proved rigorously in the early 19th century. Multiple proofs exist, using tools from complex analysis, topology, or algebra. The simplest conceptual argument uses the fact that a polynomial p(z) of degree n behaves like z^n for large |z|, and the function z^n wraps around the origin n times as z travels in a large circle — so by a topological argument (the "winding number"), p(z) must hit zero at least once. Repeated application then gives all n roots.

Beyond degree four: no general formula

For degrees 1 through 4, there exist closed-form formulas that express the roots in terms of the coefficients using only +, -, \times, \div, and n-th roots (radicals). The quadratic formula is the degree-2 case. There is a cubic formula (degree 3, much messier) and a quartic formula (degree 4, very messy).

At degree 5, something remarkable happens. A result proved independently by Abel and Galois shows that no such formula exists for the general quintic equation. It is not that nobody has found one — it is proved impossible. The operations +, -, \times, \div, and radicals are simply not powerful enough to express the roots of every fifth-degree equation.

This does not mean degree-5 equations have no roots — the Fundamental Theorem guarantees they have exactly five. It means you need other tools to find them: numerical methods, special functions, or symmetry-based techniques. The roots exist; they just resist being written down in closed form.

Symmetric functions of roots

The root-coefficient relations are instances of elementary symmetric polynomials. Given variables r_1, \ldots, r_n:

Newton's identities connect these to the power sums p_k = r_1^k + r_2^k + \cdots + r_n^k. For instance, p_2 = e_1^2 - 2e_2. This means you can compute r_1^2 + r_2^2 + \cdots + r_n^2 purely from the coefficients, without knowing the individual roots — a technique that appears frequently in competitive mathematics.

Where this leads next

Polynomial equations are the trunk from which many algebraic branches grow.