What is a polynomial?

A polynomial in one variable is a sum of monomials aₖxᵏ whose exponents are non-negative integers, such as 2x³ − x + 5. Expressions with x in a denominator or under a root are not polynomials — 1/x and √x are not of the form aₖxᵏ.

What do the remainder and factor theorems say?

The remainder theorem says the remainder on dividing W(x) by (x − a) equals W(a). The corollary — known in Polish textbooks as Bézout’s theorem — is the factor theorem: (x − a) divides W(x) exactly when W(a) = 0. A root of a polynomial and a linear factor of it are two spellings of one fact.

How do I find the integer roots of a polynomial?

If the coefficients are integers, then every integer root divides the constant term. For x³ − 2x² − 5x + 6 the constant term is 6, so the candidates are ±1, ±2, ±3, ±6 and you just substitute them. That turns an infinite search into eight checks.

How do I solve a biquadratic equation?

Substitute t = x², which turns the quartic into the quadratic at² + bt + c = 0. Solve it with the discriminant, then go back to x by taking square roots: each positive t gives two solutions ±√t, t = 0 gives one, and a negative t gives none. That is why a biquadratic can have 0, 1, 2, 3 or 4 roots.

Advanced level

Polynomials

Q: How do I divide a polynomial by a binomial x − a?

Horner’s scheme is the fastest route. Write out the coefficients (with zeros for the missing powers), copy the first one down, and compute each next entry as the previous result times a plus the next coefficient. The last number is the remainder; the earlier ones are the coefficients of the quotient, whose degree is one lower.

A polynomial is a sum of monomials — and everything you do to one is an extension of ordinary arithmetic. Degree and coefficients, adding, subtracting and multiplying, dividing by a binomial with Horner’s scheme, the remainder and factor theorems, integer roots, factoring and biquadratic equations.

Before you start

This topic builds on earlier ideas. Before you start, it's worth working through the lessons below — they'll make everything click:

Special productsSeven identities that turn multiplying out brackets into a single line — and the same seven read right to left, which is factoring. The square of a sum and of a difference, the difference of squares, the cubes and the sum and difference of cubes, proved on a drawing and used for mental arithmetic.
Quadratic equationsA quadratic equation has the unknown in the second power. Learn the general form, the discriminant, the formulas for the roots, the shortcuts for incomplete equations, Vieta formulas and how to read the solutions off a parabola.

Where this is used

Real situations where you count exactly the way this lesson teaches:

A box folded from a sheet of card
Cut squares of side x from the corners of a 30 × 20 cm sheet and fold the edges up. The volume is the polynomial V(x) = x(30 − 2x)(20 − 2x) = 4x³ − 100x² + 600x. At x = 4 cm that is 4 · 22 · 12 = 1,056 cm³, while x = 3 cm gives only 3 · 24 · 14 = 1,008 cm³ — one centimetre in the cut is almost 5% of the capacity.
Saving with a deposit every year
Three yearly deposits of 1,000 in an account paying 5% grow to 1,000(q³ + q² + q), where q = 1.05 — a cubic in the annual multiplier. Substituting: 1,000 · (1.157625 + 1.1025 + 1.05) = 3,310.13, so 310.13 on top of the 3,000 paid in.
See the lesson: Compound interest
Programming and evaluating a curve fast
A degree-five polynomial evaluated directly, with every power computed from scratch, needs 15 multiplications: 10 for the powers x² to x⁵ and 5 for the coefficients. Horner’s scheme — the same one that divides by a binomial — needs 5 multiplications and 5 additions. In a loop rendering a million points of a curve that is three times fewer multiplications per point.
Electronics and a temperature sensor
The resistance of a Pt100 sensor is the polynomial R(T) = 100(1 + 3.9083 · 10⁻³ T − 5.775 · 10⁻⁷ T²). At T = 100 °C that gives 100 · (1 + 0.39083 − 0.005775) = 138.51 Ω. Without the quadratic term it would read 139.08 Ω — half an ohm, which on this scale is an error of about 1.5 °C.
Convert it here: Temperature

All formulas

General form
$W(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$
the degree n is the highest power with a non-zero coefficient
Degree of a product
$\deg (V \cdot W) = \deg V + \deg W$
adding can drop the degree, multiplying never does
Division with remainder
$W(x) = P(x) \cdot Q(x) + R(x), \quad \deg R < \deg P$
the remainder has a lower degree than the divisor
Remainder theorem
$W(x) = (x - a) \cdot Q(x) + W(a)$
the remainder is simply the value of the polynomial at a
Factor theorem
$(x - a) \mid W(x) \iff W(a) = 0$
a root and a linear factor are the same piece of information
Integer roots
$p \mid a_0 \quad \text{for every integer root } p$
they can only be divisors of the constant term
Biquadratic equation
$ax^4 + bx^2 + c = 0 \ \xrightarrow{\ t = x^2\ } \ at^2 + bt + c = 0$
substitute t = x², then take square roots to get back to x

The expression $2x^3 - x + 5$ is a polynomial: a sum of monomials in which the variable appears only in powers with non-negative integer exponents. No $\tfrac{1}{x}$ , no $\sqrt{x}$ — and that is the whole of the restriction.

It is worth knowing straight away, because this entire lesson rests on the fact that polynomials behave exactly like numbers: you add them, multiply them, divide them with remainder and factor them. Anyone who can do long division and factor an integer already knows half of what follows.

Degree, coefficients and the constant term

W(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0

The degree is the highest power with a non-zero coefficient, and $a_0$ is the constant term — the value of the polynomial at zero. In $W(x) = x^3 - 2x^2 - 5x + 6$ the degree is $3$ and the constant term is $6$ .

Two polynomials are equal when they have the same coefficients on the same powers — agreeing at a few arguments is not enough. That principle is what every "choose the coefficients so that…" problem is built on.

Adding, subtracting and multiplying

Adding and subtracting is collecting like terms, power against power. The only trap is in the subtraction: a minus in front of a bracket flips the sign of every term inside it.

(3x^2 - x + 4) - (x^2 + 5x - 2) = 3x^2 - x + 4 - x^2 - 5x + 2 = 2x^2 - 6x + 6

Multiplying is the distributive law applied to every pair of terms. The degree of a product is always the sum of the degrees — which is where it differs from a sum, whose degree can drop when the leading terms cancel:

\deg (V \cdot W) = \deg V + \deg W

Every one of the special products is a particular case of this multiplication — and it is worth using them instead of expanding from scratch.

Dividing by a binomial: Horner’s scheme

Polynomials divide with remainder, exactly as integers do:

W(x) = P(x) \cdot Q(x) + R(x), \qquad \deg R < \deg P

When the divisor is a binomial $x - a$ , the remainder must have degree below $1$ — so it is a number. The whole computation then fits into one row of a table: Horner’s scheme. Write out the coefficients of the dividend (with zeros for missing powers), copy the first one down, and compute each next entry as the previous result times $a$ plus the next coefficient.

For $W(x) = x^3 - 2x^2 - 5x + 6$ divided by $x - 3$ (so $a = 3$ ):

	$x^3$	$x^2$	$x$	$1$
coefficients	$1$	$-2$	$-5$	$6$
$a = 3$	$1$	$1$	$-2$	$\mathbf{0}$

The last number is the remainder — here $0$ . The three before it are the coefficients of the quotient, one degree lower:

x^3 - 2x^2 - 5x + 6 = (x - 3)(x^2 + x - 2)

The remainder and factor theorems

The remainder on dividing by $x - a$ is no accident. Substitute $x = a$ into $W(x) = (x - a)Q(x) + R$ : the first term vanishes and $R = W(a)$ is left:

W(x) = (x - a) \cdot Q(x) + W(a)

From it follows the factor theorem — Bézout’s theorem in the Polish tradition — one of the most useful corollaries in all of school algebra:

(x - a) \mid W(x) \iff W(a) = 0

In other words: a root of a polynomial and a linear factor of it are one piece of information written two ways. Finding a single root immediately drops the degree of the problem by one.

Find the remainder when W(x) = 2x³ − 4x² + x − 5 is divided by x + 2.

Integer roots

The factor theorem says where the factors are — but not where to start looking. Another theorem does: if a polynomial with integer coefficients has an integer root, that root divides the constant term.

For $W(x) = x^3 - 2x^2 - 5x + 6$ the constant term is $6$ , so the candidates are $\pm 1, \pm 2, \pm 3, \pm 6$ — eight numbers instead of infinitely many. Checking in turn: $W(1) = 1 - 2 - 5 + 6 = 0$ , so $x = 1$ is a root.

The graph crosses the axis at exactly −2, 1 and 3 — the three divisors of the constant term that passed the test W(a) = 0.

Factoring

Turn the root you found into a factor, divide it out with Horner’s scheme, and carry on with a polynomial one degree lower — down to a quadratic, which factors with the discriminant or with a special product.

Factor W(x) = x³ − 2x² − 5x + 6.

Not every polynomial factors into linear pieces. A quadratic factor with a negative discriminant — $x^2 + 1$ , for instance — is irreducible over the reals and stays as it is.

Biquadratic equations

The equation $ax^4 + bx^2 + c = 0$ contains only even powers, so the substitution $t = x^2$ turns it into a quadratic:

ax^4 + bx^2 + c = 0 \ \xrightarrow{\ t = x^2\ } \ at^2 + bt + c = 0

Just remember that $t = x^2$ cannot be negative. The way back to $x$ is therefore: a positive $t$ gives two solutions $\pm\sqrt{t}$ , $t = 0$ gives one, and a negative $t$ is discarded.

Solve x⁴ − 13x² + 36 = 0.

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0

simplify (−x² + 4) + (−5x² + 7x + 6)

Common mistakes

The minus in front of a bracket — subtracting a polynomial flips the sign of every term inside, not only the first.
Missing powers in Horner’s scheme — dividing $x^3 - 7$ means writing the coefficients $1, 0, 0, -7$ . Skipping the zeros shifts the whole computation.
The wrong sign of $a$ — dividing by $x + 2$ means $a = -2$ , not $2$ . Horner’s scheme reads the root, not the number in the bracket.
Hunting for roots outside the divisors of the constant term — there are no integer roots anywhere else; if no divisor gives zero, the polynomial simply has none.
Forgetting both signs on the way back — $t = 9$ gives $x = 3$ and $x = -3$ ; keeping only the positive one loses half the solutions.
Accepting a negative $t$ — $x^2 = -4$ has no real solutions, so such a $t$ is discarded rather than square-rooted.

Formula card

Topic: Polynomials

General form
$W(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$
the degree n is the highest power with a non-zero coefficient
Degree of a product
$\deg (V \cdot W) = \deg V + \deg W$
adding can drop the degree, multiplying never does
Division with remainder
$W(x) = P(x) \cdot Q(x) + R(x), \quad \deg R < \deg P$
the remainder has a lower degree than the divisor
Remainder theorem
$W(x) = (x - a) \cdot Q(x) + W(a)$
the remainder is simply the value of the polynomial at a
Factor theorem
$(x - a) \mid W(x) \iff W(a) = 0$
a root and a linear factor are the same piece of information
Integer roots
$p \mid a_0 \quad \text{for every integer root } p$
they can only be divisors of the constant term
Biquadratic equation
$ax^4 + bx^2 + c = 0 \ \xrightarrow{\ t = x^2\ } \ at^2 + bt + c = 0$
substitute t = x², then take square roots to get back to x

The graph of W(x) = x³ − 2x² − 5x + 6. It crosses the axis at −2, 1 and 3 — exactly at its roots.

Frequently asked questions

A balance scale in perfect equilibrium rendered as a geometric form, with a straight line crossing an axis like an orbital path, against a cosmic nebula in violet and magenta

Algebra

What x really means — unknown, variable and parameter

The letter x does not always mean the same thing. Sometimes it is a number you are hunting for, sometimes a quantity that varies continuously, sometimes a blank waiting for any number at all. Here is where algebraic notation came from, why it took two thousand years to arrive, what exactly separates an unknown from a variable and a parameter, and why an inequality flips when you multiply by a negative number.

11 min