Surds, indices and polynomials in O-Level Additional Mathematics (4049): the Algebra strand foundations, from manipulating powers and roots to the remainder and factor theorems and solving cubic equations

Why these four skills come first

The Algebra strand of Additional Mathematics (SEAB 4049) opens with the manipulation skills that everything else relies on: powers, roots and polynomials. Get these secure and the rest of the syllabus, from binomial expansions to calculus, becomes far easier, because each builds on confident algebra. This module covers four outcomes that work as a set: the laws of indices, surd arithmetic and rationalising, the remainder and factor theorems, and solving polynomial equations by factorising fully.

These outcomes are not abstract drills. The laws of indices reappear in exponential and logarithmic functions and in differentiation of powers; surd skills give the exact answers the examiners ask for; and the factor theorem is the standard route into any cubic equation. This overview ties the four dot points together; each has its own worked answers and practice.

Indices: the rules for powers

The laws of indices extend multiplication of powers to every kind of exponent. The core rules are $a^m \times a^n = a^{m+n}$, $a^m \div a^n = a^{m-n}$ and $(a^m)^n = a^{mn}$, and the meanings you must hold are $a^0 = 1$, $a^{-n} = \dfrac{1}{a^n}$ and $a^{1/n} = \sqrt[n]{a}$, so that $a^{m/n} = \sqrt[n]{a^m}$. Once these are second nature you can simplify expressions and solve simple index equations by writing both sides to a common base and equating the powers.

Surds: working with exact roots

A surd is an irrational root such as $\sqrt{2}$ that we keep in exact form. The surds and rationalising outcome covers simplifying surds using $\sqrt{ab} = \sqrt{a}\,\sqrt{b}$, adding and subtracting like surds, and removing a surd from a denominator. For a single surd you multiply top and bottom by that surd; for a denominator of the form $a + \sqrt{b}$ you multiply by the conjugate $a - \sqrt{b}$, turning the denominator into the rational number $a^2 - b$.

Polynomials: the remainder and factor theorems

The remainder and factor theorems let you handle polynomials without long division. The remainder theorem gives the remainder on dividing $P(x)$ by $(x - a)$ as $P(a)$; the factor theorem is the case $P(a) = 0$, which means $(x - a)$ is a factor. In practice you test small integer values that divide the constant term until you find one giving zero, which hands you a first linear factor to extract.

Solving polynomial equations

The solving cubic and polynomial equations outcome ties the strand together. You use the factor theorem to find one root, divide out the linear factor to leave a quadratic, factorise or use the formula on that quadratic, and then apply the zero-product principle: if a product is zero, at least one factor is zero. That lists every real root.

How this module is examined

• Both papers, all questions. Paper 1 and Paper 2 (each 2 hours 15 minutes, 90 marks, 50 percent) cover the full syllabus, and you answer every question. These skills appear both alone and as steps inside larger problems.
• Watch signs and exact form. Marks are lost on careless sign errors with negative or fractional indices and on rounding when an exact surd answer is required.
• Show the method. For the factor theorem, state the value of $P(a)$ you tested and that it is zero; for full factorisation, show the division and the zero-product step, not just the roots.

Check your knowledge

Short questions on the four foundational skills. Work them with full method, then check the solutions.

1. Simplify $\dfrac{x^{1/2} \times x^{3/2}}{x^{-1}}$, giving your answer as a single power of $x$. (2 marks)
2. Rationalise the denominator of $\dfrac{4}{5 - \sqrt{3}}$. (2 marks)
3. Find the remainder when $P(x) = x^3 - 4x^2 + 2x + 1$ is divided by $(x - 2)$. (2 marks)
4. Show that $(x + 1)$ is a factor of $Q(x) = x^3 + 2x^2 - x - 2$ and factorise $Q(x)$ fully. (3 marks)