Polynomials

An algebraic expression like 2x² + 3x - 5 is called a polynomial in one variable if the variable x appears only with whole number powers (0, 1, 2, 3, ...). So expressions such as x^-1 + 2, √x + 3, or 3x^1/2 - 1 are not polynomials, because the powers there are negative or fractional.

A polynomial in x is generally written as:

p(x) = a_nxⁿ + a_n-1x^n-1 + ... + a₁x + a₀

Here a₀, a₁, ..., a_n are real numbers called coefficients, and each part such as a₂x² is called a term. The highest power of x that appears with a non-zero coefficient is called the degree of the polynomial.

Based on the number of terms, polynomials are named: a polynomial with one term is a monomial (e.g. 7x³), with two terms a binomial (e.g. x + 5), and with three terms a trinomial (e.g. x² + 2x + 1).

Based on degree, polynomials are classified as:

• Linear polynomial — degree 1, general form ax + b, a ≠ 0 (example: 3x - 7)
• Quadratic polynomial — degree 2, general form ax² + bx + c, a ≠ 0 (example: x² - 4x + 3)
• Cubic polynomial — degree 3, general form ax³ + bx² + cx + d, a ≠ 0 (example: 2x³ - x + 1)

A non-zero constant, like p(x) = 5, is a polynomial of degree 0, because 5 = 5x⁰. The zero polynomial, p(x) = 0, is a special case whose degree is not defined.

Worked Example: State whether f(x) = 5x³ - 3x² + √2 is a polynomial, and if so, find its degree and the coefficient of x².

Every power of x here (3, 2, 0) is a whole number, so f(x) is a polynomial (note that √2 is simply a real-number coefficient of x⁰; it is the power of x, not the coefficient, that must be a whole number). The highest power is 3, so its degree is 3 (a cubic polynomial). The coefficient of x² is -3.

If p(x) is a polynomial and k is a real number, then the value obtained by putting x = k in p(x) is written p(k), and is called the value of p(x) at x = k.

A real number k is called a zero of the polynomial p(x) if p(k) = 0. Finding the zero of a polynomial means finding the value(s) of x that make the polynomial equal to zero.

Worked Example 1: Find the zero of the linear polynomial p(x) = 2x - 6.

Set p(x) = 0: 2x - 6 = 0, so 2x = 6, so x = 3. Check: p(3) = 2(3) - 6 = 0. So the zero of p(x) is x = 3.

In general, every linear polynomial ax + b (a ≠ 0) has exactly one zero, given by x = -b/a.

Worked Example 2: Verify whether x = 1 and x = -1 are zeroes of p(x) = x² - 1.

p(1) = 1² - 1 = 0, so x = 1 is a zero. p(-1) = (-1)² - 1 = 1 - 1 = 0, so x = -1 is also a zero. A quadratic polynomial can have up to two zeroes.

A polynomial of degree n can have at most n zeroes. A non-zero constant polynomial has no zero, since it can never equal zero. Every real number is a zero of the zero polynomial.

When one polynomial p(x) is divided by another polynomial g(x) (not the zero polynomial), we get a quotient q(x) and a remainder r(x) such that:

p(x) = g(x) × q(x) + r(x), where r(x) = 0 or the degree of r(x) is less than the degree of g(x).

This is similar to long division of numbers — for example, 25 ÷ 7 gives quotient 3 and remainder 4, since 25 = 7 × 3 + 4.

The Remainder Theorem states: If a polynomial p(x) of degree ≥ 1 is divided by the linear polynomial (x - a), then the remainder is p(a).

This is powerful because we do not need to actually carry out long division to find the remainder — we can simply substitute x = a into p(x).

Worked Example: Find the remainder when p(x) = x³ - 3x² + 2x + 5 is divided by (x - 1).

Here g(x) = x - 1, so a = 1. By the Remainder Theorem, the remainder equals p(1).

p(1) = 1³ - 3(1)² + 2(1) + 5 = 1 - 3 + 2 + 5 = 5

So the remainder when p(x) is divided by (x - 1) is 5.

More generally, dividing p(x) by (x + a) gives remainder p(-a), and dividing by (ax - b) gives remainder p(b/a), since ax - b = 0 when x = b/a.

The Factor Theorem is a direct consequence of the Remainder Theorem. Let p(x) be a polynomial of degree ≥ 1, and let a be a real number.

1. If p(a) = 0, then (x - a) is a factor of p(x).
2. If (x - a) is a factor of p(x), then p(a) = 0.

This works because if the remainder p(a) is zero, then p(x) = (x - a) × q(x) + 0, which means (x - a) divides p(x) exactly, with no remainder.

Worked Example: Show that (x - 1) is a factor of p(x) = x³ - 6x² + 11x - 6, and hence factorise p(x) completely.

First check: p(1) = 1³ - 6(1)² + 11(1) - 6 = 1 - 6 + 11 - 6 = 0. Since p(1) = 0, (x - 1) is a factor.

Dividing p(x) by (x - 1) by long division gives quotient x² - 5x + 6, with zero remainder. So:

p(x) = (x - 1)(x² - 5x + 6)

Now factorise the quadratic x² - 5x + 6 by splitting the middle term: we need two numbers whose product is 6 and sum is -5, which are -2 and -3.

x² - 5x + 6 = x² - 2x - 3x + 6 = x(x - 2) - 3(x - 2) = (x - 2)(x - 3)

So the complete factorisation is p(x) = (x - 1)(x - 2)(x - 3).

When looking for a value of a to try, it helps to test the factors of the constant term of p(x) (here, factors of -6: ±1, ±2, ±3, ±6).

An identity is an equation that is true for every value of the variables involved. The following identities extend the familiar (x + y)² and (x - y)² identities to degree 3, and are extremely useful for expanding and factorising polynomials.

(x + y)³ = x³ + 3x²y + 3xy² + y³

(x - y)³ = x³ - 3x²y + 3xy² - y³

x³ + y³ = (x + y)(x² - xy + y²)

x³ - y³ = (x - y)(x² + xy + y²)

x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)

A special and very useful case of the last identity occurs when x + y + z = 0. Then the right-hand side becomes zero, so:

If x + y + z = 0, then x³ + y³ + z³ = 3xyz.

Worked Example 1: Expand (2x + 3y)³ using the identity for (x + y)³.

Using (a + b)³ = a³ + 3a²b + 3ab² + b³ with a = 2x, b = 3y:

(2x + 3y)³ = (2x)³ + 3(2x)²(3y) + 3(2x)(3y)² + (3y)³ = 8x³ + 36x²y + 54xy² + 27y³

Worked Example 2: Evaluate 12³ + (-7)³ + (-5)³ using the identity x³ + y³ + z³ = 3xyz (valid when x + y + z = 0).

Here x = 12, y = -7, z = -5, and x + y + z = 12 - 7 - 5 = 0. So:

12³ + (-7)³ + (-5)³ = 3 × 12 × (-7) × (-5) = 3 × 420 = 1260

This avoids the tedious work of cubing each number separately and adding them.