Mathematics (9660) | Algebra and functions

Welcome to Algebra and Functions!

Hello there! This chapter, Algebra and Functions, is arguably the most fundamental building block of advanced mathematics. Think of it as the bedrock upon which calculus, trigonometry, and every other P2 topic rests. We'll start by mastering essential manipulation skills from AS Level (P1) and then move into the exciting world of functions and advanced polynomial techniques required for A Level (P2).

Don't worry if some concepts seem familiar but tough—we will break them down step-by-step. By the end, you'll be manipulating expressions, sketching complex graphs, and solving simultaneous equations like a pro!

Section 1: Core Algebraic Manipulation (P1 Review)

1.1 Surds and Indices

Surds and indices are just different ways of dealing with powers and roots. Mastery here ensures smooth sailing in later topics.

Simplification of Surds

A surd is an irrational number that involves a root (like \(\sqrt{2}\) or \(\sqrt{5}\)). We simplify surds by looking for square number factors.

Step-by-step example: Simplify \(\sqrt{72}\)
1. Find the largest square factor of 72. (It's 36).
2. Rewrite the surd: \(\sqrt{72} = \sqrt{36 \times 2}\)
3. Separate the roots: \(\sqrt{36} \times \sqrt{2}\)
4. Simplify: \(6\sqrt{2}\)

Rationalising the Denominator

You must not leave a surd in the denominator of a fraction. This process is called rationalisation.

• If the denominator is a single surd (e.g., \( \frac{1}{\sqrt{a}} \)), multiply the top and bottom by \(\sqrt{a}\).
• If the denominator is a binomial expression with a surd (e.g., \( a+\sqrt{b} \)), multiply the top and bottom by its conjugate, which is \( a-\sqrt{b} \). This uses the difference of two squares identity: \((a-b)(a+b) = a^2 - b^2\).

Example using the conjugate: Rationalise \( \frac{1}{\sqrt{2} - 1} \)
Multiply by the conjugate, \(\sqrt{2} + 1\):
$$ \frac{1}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = \frac{\sqrt{2} + 1}{(\sqrt{2})^2 - (1)^2} = \frac{\sqrt{2} + 1}{2 - 1} = \sqrt{2} + 1 $$

Laws of Indices (for all rational exponents)

Remember these key rules for handling powers:

• Multiplication: \( a^m \times a^n = a^{m+n} \)
• Division: \( a^m \div a^n = a^{m-n} \)
• Power of a Power: \( (a^m)^n = a^{mn} \)
• Negative Power: \( a^{-n} = \frac{1}{a^n} \)
• Fractional Power (Root): \( a^{\frac{1}{n}} = \sqrt[n]{a} \)
• General Fractional Power: \( a^{\frac{m}{n}} = (\sqrt[n]{a})^m \)

1.2 Quadratic Functions

A quadratic function is an expression of the form \( ax^2 + bx + c \). Its graph is a U-shaped curve called a parabola.

Completing the Square

Completing the square converts the standard form into the vertex form, \( a(x+p)^2 + q \). This form immediately tells us the minimum (or maximum) point, or vertex, is at \((-p, q)\).

Step-by-step example: Complete the square for \( x^2 + 6x - 1 \)
1. Halve the coefficient of \(x\): \(+6 \rightarrow +3\)
2. Write the squared bracket: \((x+3)^2\)
3. Expanding this gives \(x^2 + 6x + 9\). We only want \(x^2 + 6x\), so subtract the squared term: \((x+3)^2 - 9\)
4. Reintroduce the original constant: \((x+3)^2 - 9 - 1 = (x+3)^2 - 10\)
The vertex is at \((-3, -10)\). The line of symmetry is \(x = -3\).

The Discriminant (\( \Delta \))

The discriminant, denoted \(\Delta\), is the part of the quadratic formula under the square root: \(\Delta = b^2 - 4ac\). It tells you about the nature of the roots (solutions) without solving the equation.

• If \( \mathbf{b^2 - 4ac > 0} \): There are two distinct real roots (the parabola crosses the x-axis twice).
• If \( \mathbf{b^2 - 4ac = 0} \): There is one repeated real root (the parabola touches the x-axis once at the vertex).
• If \( \mathbf{b^2 - 4ac < 0} \): There are no real roots (the parabola is entirely above or below the x-axis).

Solving Quadratic Equations

You have three main tools for solving \( ax^2 + bx + c = 0 \):

1. Factorisation: Fastest method, but only works if the roots are nice integers or simple fractions.
2. Quadratic Formula: Always works. You must learn this formula as it is NOT provided in the booklet: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$
3. Completing the Square: Useful if the question asks for the answer in surd form or if you need to find the vertex simultaneously.

1.3 Polynomials, Division, and Theorems

A polynomial is an expression built from variables and coefficients using only addition, subtraction, multiplication, and non-negative integer exponents (like \( x^3 - 5x^2 + 7x - 3 \)).

Simple Algebraic Division

When dividing a polynomial \( f(x) \) by a linear factor \( (x-a) \), you can use long division, inspection (equating coefficients), or synthetic division. The goal is to write the polynomial in the form:

$$ \frac{\text{Dividend}}{\text{Divisor}} = \text{Quotient} + \frac{\text{Remainder}}{\text{Divisor}} $$

Example: Dividing \( x^3 - x^2 - 5x + 2 \) by \( x+2 \). You will find the quotient is \( x^2 - 3x + 1 \) and the remainder is 0.

The Remainder Theorem

The Remainder Theorem is a massive shortcut! If a polynomial \( f(x) \) is divided by a linear factor \((x-a)\), the remainder is simply \( f(a) \).

Trick: If dividing by \( (x+3) \), you use \( a=-3 \). If dividing by \( (2x-1) \), you use \( a=\frac{1}{2} \). Just set the divisor equal to zero and solve for \(x\).

The Factor Theorem

The Factor Theorem is a special case of the Remainder Theorem:
If \( f(a) = 0 \), then \((x-a)\) is a factor of \( f(x) \).

This is crucial for factorising cubics. Since you only need to deal with cubics where the factor \((x-a)\) has an integer value \(a\), you can test integer factors of the constant term (e.g., test \(\pm 1, \pm 2, \pm 3, \dots\)).

Step-by-step Cubic Factorisation: Factorise \( f(x) = x^3 - 5x^2 + 7x - 3 \)
1. Test integer factors of the constant term (-3). Try \(x=1\).
2. \( f(1) = (1)^3 - 5(1)^2 + 7(1) - 3 = 1 - 5 + 7 - 3 = 0 \).
3. Since \(f(1)=0\), \((x-1)\) is a factor.
4. Divide \( f(x) \) by \((x-1)\) to find the resulting quadratic quotient (e.g., using algebraic division or inspection: \( x^3 - 5x^2 + 7x - 3 = (x-1)(x^2 - 4x + 3) \)).
5. Factorise the quadratic: \( x^2 - 4x + 3 = (x-3)(x-1) \).
6. Full factorisation: \( f(x) = (x-1)(x-1)(x-3) = (x-1)^2(x-3) \).

1.4 Simultaneous Equations and Inequalities

Solving Simultaneous Equations (Linear and Quadratic)

You will encounter problems requiring the analytical solution of one linear equation and one quadratic equation (e.g., finding the intersection of a line and a curve).

Method: Substitution

1. Make one variable (usually \(y\) or \(x\)) the subject of the linear equation.
2. Substitute this expression into the quadratic equation.
3. Solve the resulting quadratic equation for the first variable.
4. Substitute these solutions back into the linear equation (the easiest one!) to find the corresponding values of the second variable.

Geometrical Interpretation: The solutions represent the coordinates of the points where the line and the curve intersect.

• If you get 2 distinct solutions, the line cuts the curve at 2 points.
• If you get 1 repeated solution (discriminant = 0), the line is a tangent to the curve.
• If you get no real solutions (discriminant < 0), the line misses the curve entirely.

Solving Linear and Quadratic Inequalities

When solving inequalities, the rule for quadratic inequalities is different from the simple algebraic manipulation used for linear inequalities.

Linear Inequalities (Simple Algebra): Treat the inequality sign like an equals sign, but remember: If you multiply or divide by a negative number, you must reverse the inequality sign.

Quadratic Inequalities (Sketching is Key!):

Example: Solve \( 2x^2 + x \ge 6 \).
1. Rearrange to have 0 on one side: \( 2x^2 + x - 6 \ge 0 \).
2. Find the critical values by setting the expression equal to 0 and solving: \( (2x-3)(x+2) = 0 \). Critical values are \( x = \frac{3}{2} \) and \( x = -2 \).
3. Sketch the parabola \( y = 2x^2 + x - 6 \). Since \(a=2\) is positive, it's a 'happy' U-shape, crossing the x-axis at -2 and 1.5.
4. Identify the region where the curve is \(\ge 0\) (on or above the x-axis).
5. Write the solution:
$$ x \le -2 \text{ or } x \ge \frac{3}{2} $$

Section 2: Functions and Graph Transformations (P2 Extension)

Now we move into the P2 material, focusing on formal definitions of functions and how their graphs can be stretched, reflected, and shifted.

2.1 Functions, Domain, and Range

A function is a rule that maps every input value (from the domain) to exactly one output value (in the range).

• Domain: The set of possible input values (\(x\)).
• Range: The set of possible output values (\(f(x)\) or \(y\)).

Did you know? The concept of Domain and Range is crucial because not all functions are defined for all real numbers. For example, \( f(x) = \frac{1}{x-2} \) cannot accept \(x=2\), and \( g(x) = \sqrt{x} \) cannot accept negative \(x\).

Composition of Functions

When you put one function inside another, this is composition.

The notation \( fg(x) \) means "do \(g\) first, then do \(f\) to the result." So, \( fg(x) = f(g(x)) \).

Example: If \( f(x) = x^2 + 1 \) and \( g(x) = 2x \).
$$ fg(x) = f(2x) = (2x)^2 + 1 = 4x^2 + 1 $$ $$ gf(x) = g(x^2 + 1) = 2(x^2 + 1) = 2x^2 + 2 $$

Inverse Functions (\( f^{-1}(x) \))

The inverse function, \( f^{-1}(x) \), reverses the operation of \( f(x) \). If \( f(a) = b \), then \( f^{-1}(b) = a \).

Finding \( f^{-1}(x) \) step-by-step:
1. Set \( y = f(x) \).
2. Swap \( x \) and \( y \).
3. Rearrange the new equation to make \( y \) the subject.
4. Replace \( y \) with \( f^{-1}(x) \).

Graphically: The graph of \( y = f^{-1}(x) \) is a reflection of the graph of \( y = f(x) \) in the line \( y = x \).

2.2 The Modulus Function

The modulus function, written as \( |x| \), gives the absolute (positive) value of \(x\). It essentially means the distance from zero.

$$ |x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases} $$

Graphing \( y = |f(x)| \):
To sketch \( y = |f(x)| \), take the part of the original graph \( y=f(x) \) that is below the x-axis and reflect it into the positive y-region (above the x-axis).

Solving Modulus Inequalities:
The best approach is usually graphical, or by squaring both sides (ensure you don't introduce false solutions!).

Example: Solve \( |x+2| < 3|x| \).
1. Square both sides: \( (x+2)^2 < (3x)^2 \)
2. \( x^2 + 4x + 4 < 9x^2 \)
3. Rearrange into a quadratic inequality: \( 0 < 8x^2 - 4x - 4 \), or \( 2x^2 - x - 1 > 0 \).
4. Find critical values: \( (2x+1)(x-1) = 0 \). Critical values are \( x = -\frac{1}{2} \) and \( x = 1 \).
5. Sketch the positive parabola \( y = 2x^2 - x - 1 \).
6. Solution: \( x < -\frac{1}{2} \text{ or } x > 1 \).

2.3 Transformations of Graphs

You need to know the effect of four basic transformations on the graph \( y = f(x) \). Remember to treat changes inside the function (affecting \(x\)) as counter-intuitive!

Transformation	Equation	Effect on Graph	Direction
Vertical Stretch	\( y = af(x) \)	Stretched by factor \(a\) (y-coordinates multiplied by \(a\)).	y-direction
Vertical Translation (Shift)	\( y = f(x) + a \)	Translated by vector \(\begin{pmatrix} 0 \\ a \end{pmatrix}\). (Shifted up by \(a\)).	y-direction
Horizontal Translation (Shift)	\( y = f(x + a) \)	Translated by vector \(\begin{pmatrix} -a \\ 0 \end{pmatrix}\). (Shifted left by \(a\)).	x-direction (Counter-intuitive!)
Horizontal Stretch	\( y = f(ax) \)	Stretched by factor \(\frac{1}{a}\). (x-coordinates divided by \(a\)).	x-direction (Counter-intuitive!)

Memory Aid: Changes outside the brackets (affecting \(y\)) do exactly what you expect (e.g., \( +2 \) shifts up 2). Changes inside the brackets (affecting \(x\)) do the opposite (e.g., \( (x+2) \) shifts left 2).

Section 3: Rational Functions and Partial Fractions (P2 Advance)

3.1 Algebraic Division of Rational Functions

A rational function is a fraction where the numerator and denominator are polynomials (e.g., \( \frac{x^2 - 4x}{x^2 - 5x + 4} \)).

Simplification: Often, you can simplify rational expressions by factorising the numerator and denominator and cancelling common factors. $$ \frac{x^2 - 4x}{x^2 - 5x + 4} = \frac{x(x-4)}{(x-4)(x-1)} = \frac{x}{x-1} $$

If the degree (highest power) of the numerator is greater than or equal to the degree of the denominator, you must perform algebraic division first.

Example of Algebraic Division (by inspection):
$$ \frac{3x+4}{x-1} $$ Since we want \( x-1 \) to divide nicely, we manipulate the numerator:
$$ \frac{3(x-1) + 3 + 4}{x-1} = \frac{3(x-1) + 7}{x-1} = 3 + \frac{7}{x-1} $$

3.2 Partial Fractions

Partial Fractions is the reverse of adding fractions. It breaks a complex rational expression back down into simpler fractions. This is vital for later integration and series expansions.

Prerequisite: The degree of the numerator must be less than the degree of the denominator. If not, perform algebraic division first!

You need to handle two cases (irreducible quadratic factors are not tested):

Case 1: Distinct Linear Factors

If the denominator has factors \((x-a)(x-b)(x-c)\), the decomposition is:

$$ \frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b} $$

Step-by-step example: Decompose \( \frac{3x+1}{(x-1)(x+2)} \).
1. Set up the identity: \( \frac{3x+1}{(x-1)(x+2)} \equiv \frac{A}{x-1} + \frac{B}{x+2} \)
2. Clear the denominator: \( 3x+1 \equiv A(x+2) + B(x-1) \)
3. Use the Substitution Method (strategic substitution of \(x\)):
   Let \( x = 1 \): \( 3(1)+1 = A(1+2) + B(0) \implies 4 = 3A \implies A = \frac{4}{3} \).
   Let \( x = -2 \): \( 3(-2)+1 = A(0) + B(-2-1) \implies -5 = -3B \implies B = \frac{5}{3} \).
4. Final result: \( \frac{4}{3(x-1)} + \frac{5}{3(x+2)} \)

Case 2: Repeated Linear Factors

If the denominator has a repeated factor, such as \((x-a)^2\), you must include a term for every power up to the repetition count:

$$ \frac{P(x)}{(x-a)^2(x-b)} = \frac{A}{x-a} + \frac{B}{(x-a)^2} + \frac{C}{x-b} $$

Example setup for \( \frac{3 + 2x^2}{(2x+1)(x-3)^2} \): $$ \frac{A}{2x+1} + \frac{B}{x-3} + \frac{C}{(x-3)^2} $$ You clear the denominator and use substitution (for \(A\) and \(C\)) and/or equating coefficients (for \(B\)).