More about polynomials
Polynomial division, the remainder theorem and the factor theorem, and how to factorise a cubic. These skills appear regularly in the longer DSE Paper 1 questions.
What you will learn
- 4.1 Division of polynomials (long division or other methods).
- 4.2 Remainder theorem.
- 4.3 Factor theorem; use it to factorise polynomials such as x³ − a³.
- NF 4.4 H.C.F. and L.C.M. of polynomials.
- NF 4.5 Add, subtract, multiply and divide rational functions (algebraic fractions), with at most two variables.
You need first: Polynomials and factorisation, Quadratic equations in one unknown
Key terms
| English | 中文 |
|---|---|
| dividend, divisor | 被除式，除式 |
| quotient, remainder | 商式，餘式 |
| long division | 長除法 |
| remainder theorem | 餘式定理 |
| factor theorem | 因式定理 |
| H.C.F. / L.C.M. of polynomials | 最大公因式／最小公倍式 |
| rational function | 有理函數 |
Standard steps
Remainder theorem
When f(x) is divided by (x − a), the remainder is f(a). For (ax − b), the remainder is f(b/a).
- Find the value that makes the divisor zero: x − a gives a; 2x + 1 gives −1/2.
- Substitute that value into f(x).
- The result is the remainder; no long division is needed.
Why: f(x) = (x − a) × Q(x) + R. Put x = a and the first part becomes 0, leaving R.
Factor theorem
(x − a) is a factor of f(x) if and only if f(a) = 0.
- Find the value that makes the divisor zero (x − a gives a).
- Work out f(a).
- If f(a) = 0, (x − a) is a factor. If not, it is not a factor, and f(a) is the remainder.
Factorise a cubic (no calculator solver needed)
- Try x = ±1, ±2, ±3, … (factors of the constant term) until f(a) = 0.
- Then (x − a) is a factor.
- Divide f(x) by (x − a) to get a quadratic.
- Factorise the quadratic (or use the formula if it won't factorise).
Find an unknown coefficient
- "f(x) is divisible by (x − a)" means f(a) = 0.
- "Remainder r when divided by (x − a)" means f(a) = r.
- Two conditions give two equations; solve them simultaneously.
H.C.F. and L.C.M. (NF)
- Factorise each polynomial completely.
- H.C.F.: the common factors, each with its lowest power.
- L.C.M.: every factor, each with its highest power.
Worked examples
f(2) = 8 − 8 + 8 − 7 = 1. (Long division gives quotient x² + 4, remainder 1.)
f(−1) = −1 − 4 − 1 + 6 = 0, so (x + 1) is a factor
f(x) ÷ (x + 1) = x² − 5x + 6
f(x) = (x + 1)(x − 2)(x − 3)f(2) = 16 + 4k − 10 + 2 = 0
4k = −8
k = −2f(2) = 0, so x − 2 is a factor: (x − 2)(x² + 2x + 4).
x² − 1 = (x − 1)(x + 1)
x² + 3x + 2 = (x + 1)(x + 2)
H.C.F. = x + 1
L.C.M. = (x − 1)(x + 1)(x + 2)Common mistakes
- Dividing by (x + 1) and substituting x = 1 (it must be x = −1).
- Missing terms in long division: write x³ + 0x² − 8 to keep the columns.
- Stopping after one factor: the quotient quadratic may factorise further.
- Taking "remainder 0" and "divisible" as different things (they are the same).
How the DSE tests it
- A(2): find unknown coefficients from two conditions, factorise a cubic, then ask about its roots (for example, "are all the roots rational?" needs the discriminant of the quadratic factor).
- Paper 2: remainder-theorem items, often with two unknowns; NF H.C.F./L.C.M. and rational-function simplification.
Questions you might not ask in class
Why does f(a) give the remainder?
Because at x = a the "divisor times quotient" part becomes zero, so only the remainder is left.
Which numbers should I try first?
Only the factors of the constant term (with + and −). For a constant of 6: ±1, ±2, ±3, ±6.
Can I use the calculator's cubic solver?
It is useful for checking your answer. But the question usually asks you to "factorise", which needs working, so show the factor theorem.