Target Audience: Anyone who can manipulate algebra and wants to know why the rules are what they are.
Prerequisites: Tier 0 — in particular Proof Techniques, since induction proves the Binomial Theorem and the Factor Theorem is a direct proof in two lines.
This page has one destination. It starts with polynomials over the reals, where some equations have no solution, and ends at the Fundamental Theorem of Algebra, where every non-constant polynomial factors completely. Everything between is the machinery that makes that statement precise and useful.
The pivot is Section 7, where one new number is admitted and the whole subject becomes tidy. Sections 3–5 are the working tools you will use most often; the roots explorer lets you build a polynomial from its roots and watch Vieta's formulas hold.
First unit of Tier 1. It assumes the proof techniques of Tier 0 but no algebra beyond school level.
The same symbol is asking two different questions. In an equation it is a constraint to be solved; in an identity it is a claim to be proved for all values. Proving an identity means an argument covering every case — never checking a few numbers, which is proof by example.
Solving \(\sqrt{x+2} = x\) by squaring gives \(x^2 - x - 2 = 0\), so \(x = 2\) or \(x = -1\). But \(x = -1\) fails the original: \(\sqrt{1} = 1 \neq -1\).
Squaring is not reversible, so it produces a consequence, not an equivalent equation — exactly the one-way-step problem flagged for chains of equivalences. Multiplying by something that might be zero loses solutions the same way. Always check candidates against the original.
A polynomial in \(x\) is a finite sum
with coefficients \(a_i\) from a fixed number system — on this page \(\mathbb{Q}\), \(\mathbb{R}\) or \(\mathbb{C}\), all three of them fields — and non-negative integer exponents. If \(a_n \neq 0\), the degree is \(n\) and \(a_n\) is the leading coefficient. A polynomial with leading coefficient \(1\) is monic.
The exponent condition is what excludes \(1/x\) and \(\sqrt{x}\); "finite" is what excludes power series. Both restrictions are what make every theorem below true.
\(p(x) = 0\) has no non-zero leading coefficient, so its degree is left undefined (or set to \(-\infty\) by convention). This is not pedantry: it is what keeps \(\deg(pq) = \deg p + \deg q\) true without exceptions.
"Polynomial over \(\mathbb{R}\)" and "over \(\mathbb{C}\)" are different objects, and almost every question below has a different answer depending on which you mean. \(x^2 + 1\) is irreducible over \(\mathbb{R}\) and factors as \((x-i)(x+i)\) over \(\mathbb{C}\). Always say which.
For polynomials \(f\) and \(g \neq 0\), there are unique \(q\) and \(r\) with
Exactly the shape of division with remainder for integers, degree playing the role of size. Its uniqueness half is the standard uniqueness argument: assume two such pairs and force them equal.
The field matters: each step divides by the leading coefficient of \(g\). Over the integers it can fail — \(x^2\) cannot be divided by \(2x\) with a constant remainder — as Rings and Fields explains.
Dividing by the linear \(x - c\) forces \(r\) to be a constant, and substituting \(x = c\) evaluates it:
The remainder on dividing \(f\) by \(x - c\) is exactly \(f(c)\).
An immediate corollary, and a biconditional:
Both directions are one line from the Remainder Theorem — a divisor means remainder \(0\), and remainder \(0\) means \(f(c) = 0\). This single equivalence is what ties roots to factors, and everything in Section 4 depends on it.
Full proofs, with Vieta, on The Remainder Theorem, the Factor Theorem, and Vieta's Formulas.
Is \(x - 2\) a factor of \(f(x) = x^3 - 3x^2 + 4\)? Compute \(f(2) = 8 - 12 + 4 = 0\). Yes — and no long division was needed. Dividing then gives \(f(x) = (x-2)(x^2 - x - 2) = (x-2)(x-2)(x+1)\), so \(2\) is a repeated root.
\(c\) is a root (or zero) of \(f\) when \(f(c) = 0\). It has multiplicity \(m\) when \((x-c)^m \mid f\) but \((x-c)^{m+1}\) does not. A root of multiplicity \(1\) is simple.
Each root contributes a factor \(x - c\), and factors multiply degrees. More than \(n\) roots would force degree greater than \(n\).
This too needs the coefficients to form a field, or at least to have no zero divisors: working mod \(8\), \(t^2 - 1\) has the four roots \(1, 3, 5, 7\). Rings and Fields shows which step of the argument breaks.
A consequence worth having: if two polynomials of degree \(\leq n\) agree at \(n+1\) points, they are identical — their difference has too many roots to be anything but zero. That is why interpolation is unique, and why checking an identity at enough points really can prove it for polynomials, though not for functions in general.
Over \(\mathbb{R}\), \(x^2 + 1\) has degree \(2\) and no roots at all. Only over \(\mathbb{C}\) does "exactly \(n\), counted with multiplicity" become true — which is the content of Section 10.
If a monic polynomial factors completely, multiplying the factors back out relates its coefficients to its roots — without solving anything.
For monic \(p(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_0\) with roots \(r_1, \ldots, r_n\) (with multiplicity):
In one line: the coefficient of \(x^{n-k}\) is \((-1)^k e_k\), where \(e_k\) is the sum of all products of \(k\) distinct roots. Signs alternate because each factor contributes \(-r_i\), not \(r_i\).
For \(x^2 + bx + c\) with roots \(r_1, r_2\): \(r_1 + r_2 = -b\) and \(r_1 r_2 = c\). That is the mental check behind factorising \(x^2 - 5x + 6\) — find two numbers adding to \(5\) and multiplying to \(6\).
Choose a set of roots. The explorer multiplies out \(\prod (x - r_i)\), reports the expanded polynomial, and checks Vieta's formulas against the coefficients it just produced — then substitutes every root back in as an independent confirmation.
The same idea over \(\mathbb{C}\): the solutions of \(z^n = 1\). Covered in Section 9 — come back here once you have read it.
Every value shown is computed live — nothing is tabulated in advance.
Vieta answers questions about roots without finding them. "What is the sum of the squares of the roots of \(x^3 - 6x^2 + 11x - 6\)?" — use \(\sum r_i^2 = (\sum r_i)^2 - 2\sum_{i \lt j} r_i r_j = 36 - 22 = 14\), no factorising required.
Two proofs, both instructive. By induction on \(n\), where the step needs Pascal's rule \(\binom{n}{k-1} + \binom{n}{k} = \binom{n+1}{k}\). By counting: expanding the product of \(n\) brackets, a term \(a^{n-k}b^k\) arises once for each way of choosing which \(k\) brackets contribute \(b\) — and there are \(\binom{n}{k}\) such ways.
Full argument on The Binomial Theorem, and as a counting argument in the Combinatorial Proof tutorial.
Setting \(a = b = 1\) gives \(\sum_k \binom{n}{k} = 2^n\) — a set with \(n\) elements has \(2^n\) subsets.
Setting \(a = 1, b = -1\) gives \(\sum_k (-1)^k \binom{n}{k} = 0\) for \(n \geq 1\) — equally many even-sized and odd-sized subsets. Proved on The Alternating Binomial Sum.
Admit one new number \(i\) with \(i^2 = -1\). A complex number is \(z = a + bi\) with \(a, b\) real; \(a = \operatorname{Re} z\) and \(b = \operatorname{Im} z\). Note \(\operatorname{Im} z\) is a real number.
Formally \(\mathbb{C}\) is \(\mathbb{R}^2\) with a particular multiplication, so no act of faith is required — \(i\) is the pair \((0,1)\), and \(i^2 = -1\) is a computation.
Multiplication is just expanding the brackets and replacing \(i^2\) with \(-1\); there is nothing to memorise.
The conjugate is \(\bar z = a - bi\). Then \(z \bar z = a^2 + b^2\), a non-negative real number — which is what makes division work:
Conjugation respects the arithmetic: \(\overline{z+w} = \bar z + \bar w\) and \(\overline{zw} = \bar z\, \bar w\). That is exactly why real polynomials have roots in conjugate pairs.
There is no order on \(\mathbb{C}\) compatible with its arithmetic, so "\(z \lt w\)" is meaningless for non-real numbers. Inequalities apply only to real quantities such as \(|z|\). Also beware \(\sqrt{ab} = \sqrt a \sqrt b\), which fails for negatives: \(\sqrt{-1}\sqrt{-1} = -1\), not \(\sqrt{1} = 1\).
Plot \(z = a+bi\) at \((a,b)\). Its modulus is \(|z| = \sqrt{a^2+b^2}\), the distance from the origin; its argument \(\arg z\) is the angle from the positive real axis.
The argument is only defined up to adding \(2\pi\); the principal argument is the representative in \((-\pi, \pi]\). Forgetting this is the usual source of sign errors.
In polar form, multiplication multiplies moduli and adds arguments:
So multiplying by a number of modulus \(1\) is a pure rotation. Multiplying by \(i\) rotates by \(90°\) — which is what \(i^2 = -1\) means geometrically: two quarter turns face you backwards.
\(|z + w| \leq |z| + |w|\), with equality exactly when \(z\) and \(w\) point the same way. In the plane this is the ordinary statement that a side of a triangle is no longer than the other two — proved on The Triangle Inequality for Real Numbers in the one-dimensional case.
It is the "add the arguments" rule applied \(n\) times, and the standard proof is induction on \(n\), with the angle-sum identities doing the work in the step.
Take \(n = 2\): \(\cos 2\theta + i \sin 2\theta = (\cos\theta + i\sin\theta)^2 = (\cos^2\theta - \sin^2\theta) + i(2\sin\theta\cos\theta)\). Comparing real and imaginary parts gives both double-angle formulas at once.
The \(n\) solutions of \(z^n = 1\) are
equally spaced around the unit circle, forming a regular \(n\)-gon with a vertex at \(1\).
For \(n \geq 2\) they sum to zero — they are the roots of \(z^n - 1\), whose \(z^{n-1}\) coefficient is \(0\), so Vieta gives it immediately. Their product is \((-1)^{n+1}\), again by Vieta. Both are checked live in the explorer.
Over \(\mathbb{C}\) every non-zero number has exactly \(n\) distinct \(n\)th roots. The notation \(\sqrt[n]{z}\) does not name a single value without a stated convention, so writing it as though it does is where most sign errors in complex algebra begin.
Every non-constant polynomial with complex coefficients has a complex root.
Applying it repeatedly with the Factor Theorem gives the form everyone uses: a degree-\(n\) polynomial over \(\mathbb{C}\) factors completely,
and so has exactly \(n\) roots counted with multiplicity.
\(\mathbb{C}\) is algebraically closed: admitting \(i\) to solve \(x^2 + 1 = 0\) turns out to solve every polynomial equation at once. No further extension is needed — which is not at all obvious in advance, and is why the theorem is called fundamental.
Despite the name, every proof uses analysis or topology rather than algebra alone. Full argument on The Fundamental Theorem of Algebra.
If \(p\) has real coefficients and \(p(z) = 0\), then applying conjugation to the whole equation gives \(p(\bar z) = 0\) — so non-real roots come in conjugate pairs.
Pairing each with its partner gives \((z - r)(z - \bar r) = z^2 - 2(\operatorname{Re} r)z + |r|^2\), which is real. Hence every real polynomial factors over \(\mathbb{R}\) into linear and irreducible quadratic factors — and every odd-degree real polynomial has at least one real root, since its non-real roots pair off.
| Idea | The one thing to remember |
|---|---|
| Identity vs equation | An identity carries a hidden \(\forall\) and needs a proof, not a check. |
| Degree | Non-negative integer exponents, finitely many terms. The zero polynomial has no degree. |
| Remainder / Factor | Remainder on dividing by \(x-c\) is \(f(c)\); so \((x-c) \mid f \Leftrightarrow f(c)=0\). |
| Root count | At most \(n\) over any field; exactly \(n\) over \(\mathbb{C}\) with multiplicity. |
| Vieta | Coefficient of \(x^{n-k}\) is \((-1)^k e_k\). Answers root questions without finding roots. |
| Binomial Theorem | \(\binom{n}{k}\) counts which brackets give \(b\). Substituting \(1, \pm 1\) yields identities free. |
| Complex arithmetic | Expand and replace \(i^2\) by \(-1\). Divide by multiplying by the conjugate. |
| Polar form | Multiplication multiplies moduli and adds arguments — multiplying is rotating and scaling. |
| Roots of unity | A regular \(n\)-gon on the unit circle; they sum to \(0\) for \(n \geq 2\). |
| FTA | \(\mathbb{C}\) is algebraically closed. Real polynomials have non-real roots in conjugate pairs. |
Next in this series: Euclidean geometry — the other half of Tier 1, and the oldest worked example of the proof techniques from Unit 2.
| Notation | Read as | Means |
|---|---|---|
| \(\deg p\) | the degree of p | Highest power with a non-zero coefficient. Undefined for \(p = 0\). |
| \((x-c) \mid f\) | x − c divides f | Equivalent to \(f(c) = 0\) (Factor Theorem). |
| \(e_k\) | the kth elementary symmetric sum | Sum of all products of \(k\) distinct roots; coefficient of \(x^{n-k}\) is \((-1)^k e_k\). |
| \(\binom{n}{k}\) | n choose k | \(\dfrac{n!}{k!(n-k)!}\) — the number of \(k\)-subsets of an \(n\)-set. |
| \(i\) | the imaginary unit | \(i^2 = -1\). Geometrically, a quarter turn. |
| \(\operatorname{Re} z,\ \operatorname{Im} z\) | real and imaginary parts | Both are real numbers: for \(z = a+bi\), they are \(a\) and \(b\). |
| \(\bar z\) | the conjugate of z | \(a - bi\). Satisfies \(z\bar z = |z|^2\), and respects sums and products. |
| \(|z|\) | the modulus of z | \(\sqrt{a^2+b^2}\) — distance from the origin. |
| \(\arg z\) | the argument of z | Angle from the positive real axis, defined up to \(2\pi\); principal value in \((-\pi, \pi]\). |
| \(\omega_k\) | an nth root of unity | \(\cos\frac{2\pi k}{n} + i\sin\frac{2\pi k}{n}\); the \(n\) of them sum to \(0\) for \(n \geq 2\). |