1. Overview
Polynomials are a fundamental topic in IGCSE Additional Mathematics. You'll need to master factorising, finding remainders, and solving polynomial equations (especially cubics) to succeed in coordinate geometry, calculus, and other advanced topics. This section covers the Remainder and Factor Theorems, algebraic long division, and how to use these tools to solve cubic equations efficiently. Expect to see these skills tested in both Paper 1 (without a calculator) and Paper 2.
Key Definitions
- Polynomial: An expression of the form $P(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0$, where $n$ is a non-negative integer.
- Degree: The highest power of the variable $x$ in the polynomial.
- Remainder Theorem: A rule used to find the remainder when a polynomial is divided by a linear factor $(ax - b)$ without using long division.
- Factor Theorem: A specific case of the Remainder Theorem where the remainder is zero, indicating that $(ax - b)$ is a factor.
- Root (or Zero): The value of $x$ for which $P(x) = 0$.
Core Content
The Remainder Theorem
When a polynomial $P(x)$ is divided by $(ax - b)$, the remainder is given by $P\left(\frac{b}{a}\right)$.
- If dividing by $(x - c)$, the remainder is $P(c)$.
- If dividing by $(x + c)$, the remainder is $P(-c)$.
The Factor Theorem
$(ax - b)$ is a factor of $P(x)$ if and only if $P\left(\frac{b}{a}\right) = 0$. This is the most powerful tool for breaking down cubic equations into manageable parts.
Factoring Cubic Polynomials
To factorize a cubic $P(x) = ax^3 + bx^2 + cx + d$:
- Find the first linear factor: Use the Factor Theorem by testing factors of the constant term $d$ (e.g., $\pm 1, \pm 2$).
- Reduce to quadratic: Once a factor $(x - k)$ is found, divide $P(x)$ by $(x - k)$ using algebraic long division or the method of equating coefficients to find the quadratic quotient.
- Factorize the quadratic: Use standard methods (inspection, completing the square, or the quadratic formula) to find the remaining factors.
Worked Example 1 — Remainder Theorem
Find the remainder when $P(x) = 2x^3 - 5x^2 + x - 7$ is divided by $(x - 3)$.
Step 1: Identify the value to substitute. By the Remainder Theorem, $x - 3 = 0 \implies x = 3$.
Step 2: Substitute into $P(x)$. $P(3) = 2(3)^3 - 5(3)^2 + (3) - 7$
Step 3: Evaluate. $P(3) = 2(27) - 5(9) + 3 - 7$ $P(3) = 54 - 45 + 3 - 7$ $P(3) = 5$
Result: The remainder is $\boxed{5}$.
Worked Example 2 — Fully Factorizing a Cubic
Fully factorize $f(x) = x^3 - 6x^2 + 11x - 6$.
Step 1: Find a linear factor using the Factor Theorem. Try $f(1)$: $1^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0$. Since $f(1) = 0$, $(x - 1)$ is a factor.
Step 2: Use algebraic long division (or inspection) to find the quadratic. $x^3 - 6x^2 + 11x - 6 = (x - 1)(ax^2 + bx + c)$ By observation:
- $x \times ax^2 = x^3 \implies a = 1$
- $-1 \times c = -6 \implies c = 6$ Comparing $x^2$ terms: $-1x^2 + bx^2 = -6x^2 \implies b = -5$ So, $f(x) = (x - 1)(x^2 - 5x + 6)$.
Step 3: Factorize the quadratic. $x^2 - 5x + 6 = (x - 2)(x - 3)$
Final Answer: $f(x) = (x - 1)(x - 2)(x - 3)$
Worked Example 3 — Solving Equations and the Discriminant
Solve $x^3 - 3x + 2 = 0$.
Step 1: Find a root. Try $x = 1$: $1^3 - 3(1) + 2 = 0$. Thus $(x - 1)$ is a factor.
Step 2: Division. Divide $(x^3 - 3x + 2)$ by $(x - 1)$ to get $(x^2 + x - 2)$. Equation becomes: $(x - 1)(x^2 + x - 2) = 0$.
Step 3: Solve the quadratic. $(x - 1)(x + 2)(x - 1) = 0$ $(x - 1)^2(x + 2) = 0$
Result: $x = 1$ (repeated root) or $x = -2$.
Worked Example 4 — Using the Remainder and Factor Theorems Together
The polynomial $P(x) = 2x^3 - 3x^2 - 11x + 6$.
(a) Show that $(x - 3)$ is a factor of $P(x)$. (b) Hence factorise $P(x)$ completely.
Step 1: Apply the Factor Theorem for part (a). $P(3) = 2(3)^3 - 3(3)^2 - 11(3) + 6 = 54 - 27 - 33 + 6 = 0$ Since $P(3) = 0$, $(x - 3)$ is a factor of $P(x)$. ✓
Step 2: Divide $P(x)$ by $(x - 3)$ using algebraic long division (or comparing coefficients). $2x^3 - 3x^2 - 11x + 6 = (x - 3)(2x^2 + 3x - 2)$
Check by expanding: $(x-3)(2x^2 + 3x - 2) = 2x^3 + 3x^2 - 2x - 6x^2 - 9x + 6 = 2x^3 - 3x^2 - 11x + 6$ ✓
Step 3: Factorise the quadratic $2x^2 + 3x - 2$. $2x^2 + 3x - 2 = (2x - 1)(x + 2)$
Step 4: Write the complete factorisation. $P(x) = (x - 3)(2x - 1)(x + 2)$