Polynomial Manipulation

This lesson covers polynomial manipulation as required by the AQA A-Level Mathematics specification (7357). Polynomials are expressions involving powers of a variable with non-negative integer exponents. At A-Level, you must be able to add, subtract, multiply, and divide polynomials, apply the factor theorem and remainder theorem, and use algebraic long division and nested multiplication (Horner's method). These skills underpin much of pure mathematics and are essential for solving equations, sketching curves, and simplifying algebraic expressions.

What Is a Polynomial?

A polynomial in x is an expression of the form:

aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀

where n is a non-negative integer (the degree of the polynomial) and aₙ, aₙ₋₁, ..., a₀ are constants called coefficients. The term aₙxⁿ (where aₙ ≠ 0) is the leading term and aₙ is the leading coefficient.

Term	Meaning
Degree	The highest power of x with a non-zero coefficient
Monic polynomial	A polynomial whose leading coefficient is 1
Constant term	The term a₀ (the value when x = 0)

Examples:

3x⁴ − 2x² + 7 is a polynomial of degree 4
x³ + x − 1 is a monic polynomial of degree 3
5 is a polynomial of degree 0 (a constant)

Adding and Multiplying Polynomials

Addition and Subtraction

Combine like terms (terms with the same power of x).

Example: (3x³ − 2x² + 4x − 1) + (x³ + 5x² − 3x + 7)

= 4x³ + 3x² + x + 6

Multiplication

Multiply every term in one polynomial by every term in the other, then collect like terms.

Example: (2x + 3)(x² − x + 4)

= 2x(x² − x + 4) + 3(x² − x + 4)
= 2x³ − 2x² + 8x + 3x² − 3x + 12
= 2x³ + x² + 5x + 12

Exam Tip: When multiplying polynomials, work systematically. Multiply each term in the first bracket by every term in the second bracket. It can help to use a grid method for more complex products.

Algebraic Long Division

To divide a polynomial f(x) by a divisor d(x), we use algebraic long division. This produces a quotient q(x) and a remainder r(x) such that:

f(x) = d(x) × q(x) + r(x)

where the degree of r(x) is less than the degree of d(x).

Method

Divide the leading term of f(x) by the leading term of d(x) — this gives the first term of the quotient.
Multiply d(x) by that term and subtract from f(x).
Repeat with the resulting polynomial until the degree of the remainder is less than the degree of d(x).

Example: Divide x³ + 2x² − 5x − 6 by (x + 3).

         x² − x − 2
    ___________________
x+3 | x³ + 2x² − 5x − 6
      x³ + 3x²
      ---------
          −x² − 5x
          −x² − 3x
          ---------
               −2x − 6
               −2x − 6
               --------
                    0

So x³ + 2x² − 5x − 6 = (x + 3)(x² − x − 2) with remainder 0.

Example: Divide 2x³ − 3x² + x + 7 by (x − 2).

          2x² + x + 3
    ___________________
x−2 | 2x³ − 3x² + x + 7
      2x³ − 4x²
      ----------
            x² + x
            x² − 2x
            --------
                3x + 7
                3x − 6
                ------
                    13

So 2x³ − 3x² + x + 7 = (x − 2)(2x² + x + 3) + 13.

The Factor Theorem

The factor theorem states:

If f(a) = 0, then (x − a) is a factor of f(x).

Conversely, if (x − a) is a factor of f(x), then f(a) = 0.

This is a special case of the remainder theorem (see below) when the remainder is zero.

Example: Show that (x − 2) is a factor of f(x) = x³ − 6x² + 11x − 6.

f(2) = 8 − 24 + 22 − 6 = 0

Since f(2) = 0, (x − 2) is a factor of f(x) by the factor theorem. ∎

Example: Factorise f(x) = x³ − 6x² + 11x − 6 completely.

Since f(2) = 0, (x − 2) is a factor. Dividing:

x³ − 6x² + 11x − 6 = (x − 2)(x² − 4x + 3) = (x − 2)(x − 1)(x − 3)

Exam Tip: To find factors, try substituting x = ±1, ±2, ±3, ... (factors of the constant term divided by factors of the leading coefficient). Once you find one factor, use algebraic long division or inspection to find the remaining quadratic, then factorise or use the formula.

The Remainder Theorem

The remainder theorem states:

When a polynomial f(x) is divided by (x − a), the remainder is f(a).

More generally, when f(x) is divided by (ax − b), the remainder is f(b/a).

Example: Find the remainder when f(x) = 2x³ − x² + 3x − 4 is divided by (x − 1).

f(1) = 2 − 1 + 3 − 4 = 0

The remainder is 0 — so (x − 1) is actually a factor.

Example: Find the remainder when f(x) = x⁴ + 2x³ − x + 5 is divided by (x + 2).

f(−2) = 16 − 16 + 2 + 5 = 7

The remainder is 7.

Example: Find the remainder when f(x) = 4x³ − 2x + 1 is divided by (2x − 1).

f(1/2) = 4(1/8) − 2(1/2) + 1 = 1/2 − 1 + 1 = 1/2

The remainder is 1/2.

Nested Multiplication (Horner's Method)

Nested multiplication (also called Horner's method or Horner's scheme) is an efficient way to evaluate a polynomial at a given value. It also provides a compact method for polynomial division by a linear factor.

Evaluating a Polynomial

Rewrite the polynomial in nested form by factoring out x repeatedly:

f(x) = 2x³ − 3x² + x − 5
     = ((2x − 3)x + 1)x − 5

To evaluate f(4):

Start with 2
2 × 4 = 8;   8 − 3 = 5
5 × 4 = 20;  20 + 1 = 21
21 × 4 = 84; 84 − 5 = 79

So f(4) = 79.

Synthetic Division (Tabular Horner's Method)

Synthetic division is a shorthand for dividing by (x − a). Write down the coefficients and use a table.

Example: Divide 2x³ + 3x² − 8x + 3 by (x − 1) using synthetic division.

   | 2    3    −8    3
 1 |      2     5   −3
   |____________________
     2    5    −3    0

Reading off: quotient is 2x² + 5x − 3, remainder is 0.

So 2x³ + 3x² − 8x + 3 = (x − 1)(2x² + 5x − 3).

We can further factorise: 2x² + 5x − 3 = (2x − 1)(x + 3).

Therefore 2x³ + 3x² − 8x + 3 = (x − 1)(2x − 1)(x + 3).

Finding Unknown Coefficients

The factor and remainder theorems can be used to find unknown coefficients in a polynomial.

Example: Given that (x + 2) is a factor of f(x) = x³ + kx² − x + 6, find k.

By the factor theorem, f(−2) = 0:

(−2)³ + k(−2)² − (−2) + 6 = 0
−8 + 4k + 2 + 6 = 0
4k = 0
k = 0

Example: When f(x) = 2x³ + ax² + bx − 3 is divided by (x − 1) the remainder is 4, and (x + 1) is a factor. Find a and b.

f(1) = 2 + a + b − 3 = 4   →   a + b = 5   ... (1)
f(−1) = −2 + a − b − 3 = 0  →   a − b = 5   ... (2)

Adding (1) and (2): 2a = 10, so a = 5. From (1): b = 0.

Summary

A polynomial of degree n has leading term aₙxⁿ.
Algebraic long division divides f(x) by d(x) to give a quotient and remainder.
The factor theorem: (x − a) is a factor of f(x) if and only if f(a) = 0.
The remainder theorem: when f(x) is divided by (x − a), the remainder is f(a).
Nested multiplication (Horner's method) evaluates polynomials efficiently and provides a compact route to synthetic division.
Use the factor and remainder theorems to find unknown coefficients by forming and solving simultaneous equations.

Exam Tip: When factorising a cubic, always start by testing small integer values (x = ±1, ±2, ±3) to find a root using the factor theorem. Once you have one factor, divide to find the quadratic factor. Always check your factorisation by expanding. In the exam, show the substitution f(a) = 0 clearly and state the factor theorem by name.