Learn first: Powers and their properties
Polynomials are the “raw material” of algebra: equations, functions and expressions are built from them. Quadratic trinomials, cubic equations, finding derivatives — all of these require working with polynomials. In this lesson we start with monomials and polynomials, then cover the operations, polynomial division, the remainder theorem (Bézout’s theorem), Horner’s scheme and finding roots — the key to higher-degree equations.
A monomial is a product of a number and powers of variables: −3x²y, 5a, 7. The number is the coefficient (−3). The degree of a monomial is the sum of the exponents of its variables: −3x²y has degree 2 + 1 = 3.
Like terms differ only in their coefficient: 4x²y and −x²y. When added, their coefficients add: 4x²y − x²y = 3x²y.
A polynomial is a sum of monomials. In standard form like terms are collected and the powers are written in decreasing order: P(x) = 2x³ − x + 5.
Substituting numbers into a polynomial gives very useful results:
The sum of the coefficients equals P(1)
P(−1) equals the sum of the coefficients of even powers minus the sum of the coefficients of odd powers
For example, the sum of the coefficients of (x² − 3x + 1)⁵ without expanding: P(1) = (1 − 3 + 1)⁵ = −1.
Addition and subtraction: remove the brackets and collect like terms. If there is a minus before a bracket, the signs of all the terms inside change:
(x² − x + 3) − (2x² − 4x − 1) = x² − x + 3 − 2x² + 4x + 1 = −x² + 3x + 4.
Multiplication: multiply every term of the first polynomial by every term of the second, then collect like terms:
(2x − 3)(x + 4) = 2x² + 8x − 3x − 12 = 2x² + 5x − 12.
When multiplying, the leading terms multiply, so:
deg(P·Q) = deg P + deg Q
deg(P + Q) ≤ max(deg P, deg Q) (the leading terms may cancel)
The special product formulas speed up multiplication:
(a ± b)² = a² ± 2ab + b², a² − b² = (a − b)(a + b)
(a ± b)³ = a³ ± 3a²b + 3ab² ± b³, a³ ± b³ = (a ± b)(a² ∓ ab + b²)
For example, (x + 2)³ = x³ + 6x² + 12x + 8.
Find the degree, the leading coefficient and the constant term of (x² + 1)³(x − 2)².
Like numbers, polynomials can be divided with remainder. For any P(x) and non-zero Q(x) there are a unique quotient S(x) and remainder R(x):
P(x) = Q(x)·S(x) + R(x), deg R < deg Q
Division is done as long division: divide the leading term of the dividend by the leading term of the divisor, multiply the divisor by the result and subtract, and repeat until the degree of the remainder is less than that of the divisor. If R(x) = 0, then P is exactly divisible by Q.
Divide x³ − 2x² + 4 by x − 2.
When dividing by x − a, the remainder is a number (degree 0). Substituting x = a into P(x) = (x − a)S(x) + R, the first term vanishes and we are left with P(a) = R. This is Bézout’s theorem:
The remainder when P(x) is divided by x − a equals P(a)
Corollary (factor theorem): P(a) = 0 ⇔ P(x) is divisible by (x − a), i.e. a is a root
The polynomial x³ + ax − 6 is divisible by x − 2. Find a.
Horner’s scheme (synthetic division) is a fast way to divide by x − a. Write the coefficients in a row (0 for missing powers). Bring the first coefficient down; then each time multiply the latest result by a and add it to the next coefficient. The last number is the remainder; the earlier ones are the coefficients of the quotient.
Divide 2x³ − 3x² + 5 by x − 2 using Horner’s scheme.
The integer roots of a polynomial with integer coefficients are among the divisors of the constant term. Reason: if a is a root, then in P(a) = 0 every term except the constant is divisible by a, so the constant term is divisible by a as well.
Algorithm:
A useful observation: if the coefficients add up to 0, then x = 1 is a root; if the sums of the coefficients of the even and odd powers are equal, then x = −1 is a root.
Solve x³ − 6x² + 11x − 6 = 0.