Welcome to AS 1 Pure Mathematics: Algebra and Functions
Welcome to your study notes for Algebra and Functions, the foundational chapter of CCEA AS 1: Pure Mathematics (Subject Code: 2210). Pure algebra makes up the core toolkit you will rely on across all of AS and A2 Mathematics. Unit AS 1 is assessed by a 1 hour 45 minute paper carrying 100 marks (worth 60% of your AS Level and 24% of your full A Level).
Don't worry if algebra has felt overwhelming in the past. We will break down every rule, skill, and exam technique step by step, using clear examples, practical memory tricks, and direct call-outs of common examiner traps so you can secure top marks.
---1. Laws of Indices and Surds
Laws of Indices
Indices (powers or exponents) tell us how many times a base number is multiplied by itself. To manipulate complex algebraic fractions effortlessly, you must know these core rules inside out:
1. Multiplication Rule: \(x^a \times x^b = x^{a+b}\)
Example: \(x^4 \times x^3 = x^{4+3} = x^7\)
2. Division Rule: \(\frac{x^a}{x^b} = x^{a-b}\)
Example: \(\frac{x^7}{x^2} = x^{7-2} = x^5\)
3. Power of a Power Rule: \((x^a)^b = x^{ab}\)
Example: \((x^3)^4 = x^{12}\)
4. Zero Index Rule: \(x^0 = 1\) (where \(x \neq 0\))
Example: \(5^0 = 1\), \((3x)^0 = 1\), but note that \(3x^0 = 3(1) = 3\).
5. Negative Index Rule: \(x^{-n} = \frac{1}{x^n}\)
Example: \(x^{-3} = \frac{1}{x^3}\), and \(\left(\frac{2}{3}\right)^{-2} = \left(\frac{3}{2}\right)^2 = \frac{9}{4}\)
6. Fractional Indices (Roots and Powers): \(x^{1/n} = \sqrt[n]{x}\) and \(x^{m/n} = \sqrt[n]{x^m} = (\sqrt[n]{x})^m\)
Memory Tip: Think of a flower: the Power is on Top (in the sun), and the Root is on the Bottom (in the ground).
Example: Evaluate \(27^{2/3}\):
First take the cube root: \(\sqrt[3]{27} = 3\).
Then square the result: \(3^2 = 9\). Thus, \(27^{2/3} = 9\).
Working with Surds
A surd is an irrational root that cannot be written as an exact fraction (e.g., \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{5}\)). Working in surd form preserves exact values.
Fundamental Surd Properties:
• \(\sqrt{ab} = \sqrt{a}\sqrt{b}\)
• \(\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}\)
• \(\sqrt{a} \times \sqrt{a} = a\)
Simplifying Surds: Identify the largest square number factor inside the square root.
\(\sqrt{72} = \sqrt{36 \times 2} = \sqrt{36}\sqrt{2} = 6\sqrt{2}\)
Rationalising the Denominator
In mathematics, it is standard convention not to leave an irrational root in the denominator of a fraction.
Case 1: Monomial Denominator (\(\frac{k}{\sqrt{a}}\))
Multiply numerator and denominator by \(\sqrt{a}\):
\(\frac{6}{\sqrt{3}} = \frac{6 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}\)
Case 2: Binomial Denominator (\(a \pm \sqrt{b}\) or \(\sqrt{a} \pm \sqrt{b}\))
Multiply numerator and denominator by the algebraic conjugate (change the middle sign) to exploit the difference of two squares: \((u+v)(u-v) = u^2 - v^2\).
Step-by-Step Example: Rationalise \(\frac{4}{3 - \sqrt{5}}\)
• Conjugate of \(3 - \sqrt{5}\) is \(3 + \sqrt{5}\).
• Multiply top and bottom: \(\frac{4(3 + \sqrt{5})}{(3 - \sqrt{5})(3 + \sqrt{5})}\)
• Expand denominator: \(3^2 - (\sqrt{5})^2 = 9 - 5 = 4\)
• Simplify: \(\frac{4(3 + \sqrt{5})}{4} = 3 + \sqrt{5}\)
Key Takeaway for Indices & Surds: Never leave surds unsimplified or fractional/negative powers in intermediate forms when simplifying exact algebraic answers. Always multiply by the exact conjugate to clear square roots from denominators.
---2. Quadratic Functions, Equations, and Inequalities
General Form and Methods of Solution
A quadratic function has the general form \(f(x) = ax^2 + bx + c\), where \(a \neq 0\).
To solve \(ax^2 + bx + c = 0\), we have three primary methods:
1. Factorisation: Useful when roots are rational.
Example: \(2x^2 - 5x - 3 = 0 \implies (2x + 1)(x - 3) = 0 \implies x = -\frac{1}{2} \text{ or } x = 3\)
2. Completing the Square: Essential for finding vertices and maximum/minimum values.
For \(x^2 + bx + c\): write as \(\left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c\).
When \(a \neq 1\), factor out \(a\) from the \(x\)-terms first:
\(f(x) = a\left[x^2 + \frac{b}{a}x\right] + c = a\left(x + \frac{b}{2a}\right)^2 + \left(c - \frac{b^2}{4a}\right) = a(x + p)^2 + q\)
• Turning Point (Vertex): \((-p, q)\)
• Axis of Symmetry: \(x = -p\)
• If \(a > 0\), the curve is U-shaped and \(q\) is the minimum value.
• If \(a < 0\), the curve is \(\cap\)-shaped and \(q\) is the maximum value.
3. The Quadratic Formula: Works for any quadratic equation:
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
The Discriminant (\(\Delta = b^2 - 4ac\))
The discriminant, \(\Delta = b^2 - 4ac\), determines the nature and number of real roots of \(ax^2 + bx + c = 0\):
• \(\Delta > 0\): Two distinct real roots (graph intersects the \(x\)-axis at two distinct points).
• \(\Delta = 0\): One repeated real root / two equal real roots (graph touches the \(x\)-axis; the \(x\)-axis is a tangent).
• \(\Delta < 0\): No real roots (graph does not cross or touch the \(x\)-axis).
• \(\Delta \geq 0\): Real roots exist (includes both distinct and repeated roots).
Solving Quadratic Inequalities
Solving a quadratic inequality requires finding the critical values and sketching the parabola.
Step 1: Rearrange all terms to one side so that you have \(ax^2 + bx + c > 0\) (or \(< 0\), \(\geq 0\), \(\leq 0\)).
Step 2: Set the quadratic equal to zero and solve to find the critical values (the roots \(x_1\) and \(x_2\), where \(x_1 < x_2\)).
Step 3: Sketch the parabola (a positive \(x^2\) coefficient gives a U-shape).
Step 4: Read off the regions from your sketch:
• If \(ax^2 + bx + c > 0\) (strictly above the \(x\)-axis):
Two separate regions: \(x < x_1 \text{ or } x > x_2\)
• If \(ax^2 + bx + c < 0\) (strictly below the \(x\)-axis):
One single continuous region: \(x_1 < x < x_2\)
Common Error Alert: Never write two disjoint sets as a single double inequality. Writing "\(3 > x > 7\)" is mathematically incorrect and loses full marks. Always write: "\(x < 3 \text{ or } x > 7\)".
Key Takeaway for Quadratics: Always sketch the parabola when solving inequalities. Use \(\Delta = b^2 - 4ac\) to classify roots without having to solve the full equation.
---3. Simultaneous Equations
Linear and Non-Linear Systems
At AS Level, you are required to solve systems containing one linear equation and one quadratic/non-linear equation.
Standard Algebraic Method:
Step 1: Rearrange the linear equation to make one variable the subject (e.g., \(y = mx + c\) or \(x = \dots\)).
Step 2: Substitute this expression into the non-linear equation.
Step 3: Simplify and rearrange into standard quadratic form: \(Ax^2 + Bx + C = 0\).
Step 4: Solve the quadratic equation for \(x\).
Step 5: Substitute each \(x\)-value back into the linear equation to obtain the corresponding \(y\)-values.
Step 6: Present solutions clearly as coordinate pairs: \((x_1, y_1)\) and \((x_2, y_2)\).
Geometrical Interpretation & Line-Curve Intersections
The algebraic solutions represent the exact points of intersection between the line and the curve.
When you substitute a linear equation into a curve's equation to form \(Ax^2 + Bx + C = 0\), the discriminant \(\Delta = B^2 - 4AC\) reveals the geometric relationship:
• \(\Delta > 0\): The line intersects the curve at two distinct points (the line is a secant/chord).
• \(\Delta = 0\): The line touches the curve at exactly one point (the line is a tangent).
• \(\Delta < 0\): The line does not intersect the curve at all.
Example Question: Show that the line \(y = 2x - 5\) is a tangent to the curve \(y = x^2 - 4x + 4\).
Set equations equal: \(2x - 5 = x^2 - 4x + 4\)
Rearrange: \(x^2 - 6x + 9 = 0\)
Calculate discriminant: \(\Delta = (-6)^2 - 4(1)(9) = 36 - 36 = 0\)
Conclusion: Since \(\Delta = 0\), there is exactly one repeated point of contact, proving the line is a tangent.
Key Takeaway for Simultaneous Equations: Always substitute into the linear equation to find the second coordinate to avoid introducing false pairs.
---4. Polynomials: Factor and Remainder Theorems
The Remainder Theorem
When a polynomial \(P(x)\) is divided by a linear divisor \((ax - b)\), the division can be expressed as:
\(P(x) = (ax - b)Q(x) + R\)
where \(Q(x)\) is the quotient and \(R\) is the constant remainder.
If we substitute \(x = \frac{b}{a}\):
\(P\left(\frac{b}{a}\right) = \left(a\left(\frac{b}{a}\right) - b\right)Q\left(\frac{b}{a}\right) + R = 0 \times Q\left(\frac{b}{a}\right) + R = R\)
The Remainder Theorem: If a polynomial \(P(x)\) is divided by \((ax - b)\), the remainder is \(R = P\left(\frac{b}{a}\right)\).
Example: Find the remainder when \(P(x) = 2x^3 - 3x^2 + 4x - 5\) is divided by \((x - 2)\).
Here, \(x - 2 = 0 \implies x = 2\).
\(R = P(2) = 2(2)^3 - 3(2)^2 + 4(2) - 5 = 2(8) - 3(4) + 8 - 5 = 16 - 12 + 8 - 5 = 7\).
The Factor Theorem
The Factor Theorem is a special case of the Remainder Theorem where the remainder is zero (\(R = 0\)).
The Factor Theorem: \((ax - b)\) is a factor of a polynomial \(P(x)\) if and only if \(P\left(\frac{b}{a}\right) = 0\).
Crucial Sign Tip:
• To test \((x - c)\), calculate \(P(c)\).
• To test \((x + c)\), calculate \(P(-c)\).
Factorising Cubics Completely
To factorise a cubic polynomial \(P(x) = ax^3 + bx^2 + cx + d\):
Step 1: Use the Factor Theorem with integer trial values (\(\pm 1, \pm 2, \dots\)) until you find a value where \(P(k) = 0\). This identifies \((x - k)\) as a factor.
Step 2: Divide \(P(x)\) by \((x - k)\) using algebraic long division (or equating coefficients) to obtain a quadratic quotient \(Q(x)\):
\(P(x) = (x - k)(Ax^2 + Bx + C)\)
Step 3: Factorise the quadratic expression \(Ax^2 + Bx + C\) into two linear brackets, if possible.
Step 4: Write the final answer as a product of three linear factors: \((x - k)(px + q)(rx + s)\).
Key Takeaway for Polynomials: When using the factor theorem, always show the explicit substitution step (e.g., \(P(2) = 2(2)^3 - \dots = 0\)) before writing "therefore \((x - 2)\) is a factor" to guarantee method marks.
---5. Graphs and Function Transformations
Standard Curve Shapes to Recognise
1. Linear: \(y = mx + c\) (straight line with gradient \(m\) and \(y\)-intercept \(c\)).
2. Quadratic: \(y = ax^2 + bx + c\) (parabola; U-shaped for \(a > 0\), \(\cap\)-shaped for \(a < 0\)).
3. Cubic: \(y = ax^3 + bx^2 + cx + d\) (S-shaped curve with up to two turning points).
4. Reciprocal Functions:
• \(y = \frac{k}{x}\): Hyperbola with asymptotes at \(x = 0\) (\(y\)-axis) and \(y = 0\) (\(x\)-axis).
• \(y = \frac{k}{x^2}\): Symmetrical about the \(y\)-axis; both branches stay on the same side of the \(x\)-axis (positive if \(k > 0\)), with asymptotes at \(x = 0\) and \(y = 0\).
Essential Graph Sketching Rules: Every sketch must clearly label all intersections with coordinate axes, the coordinates of any turning points, and the equations of any asymptotes.
Function Transformations of \(y = f(x)\)
Transformations can be grouped into vertical changes (affecting \(y\)) and horizontal changes (affecting \(x\)).
1. Vertical Transformations (Outside the bracket — behaves normally)
• Vertical Translation: \(y = f(x) + a\)
Shift upwards by \(a\) units. Vector: \(\begin{pmatrix} 0 \\ a \end{pmatrix}\).
Coordinate change: \((x, y) \to (x, y + a)\)
• Vertical Stretch: \(y = a\,f(x)\)
Stretch by scale factor \(a\) parallel to the \(y\)-axis.
Coordinate change: \((x, y) \to (x, ay)\)
• Reflection in the \(x\)-axis: \(y = -f(x)\)
Coordinate change: \((x, y) \to (x, -y)\)
2. Horizontal Transformations (Inside the bracket — behaves counter-intuitively)
• Horizontal Translation: \(y = f(x + a)\)
Shift left by \(a\) units. Vector: \(\begin{pmatrix} -a \\ 0 \end{pmatrix}\).
Notice: \(y = f(x - c)\) shifts to the right by \(c\) units.
Coordinate change: \((x, y) \to (x - a, y)\)
• Horizontal Stretch: \(y = f(ax)\)
Stretch by scale factor \(\frac{1}{a}\) parallel to the \(x\)-axis.
Coordinate change: \((x, y) \to \left(\frac{x}{a}, y\right)\)
• Reflection in the \(y\)-axis: \(y = f(-x)\)
Coordinate change: \((x, y) \to (-x, y)\)
Summary Table for a Given Point \((p, q)\):
• \(y = f(x) + a \implies (p, q + a)\)
• \(y = f(x + a) \implies (p - a, q)\)
• \(y = a\,f(x) \implies (p, aq)\)
• \(y = f(ax) \implies \left(\frac{p}{a}, q\right)\)
• \(y = -f(x) \implies (p, -q)\)
• \(y = f(-x) \implies (-p, q)\)
Key Takeaway for Transformations: Anything inside the bracket \(f(\dots)\) operates horizontally on \(x\) in the opposite way you might expect (divide by scale factor, shift opposite to sign). Anything outside operates vertically on \(y\) exactly as written.
---6. CCEA Examination Rules & Examiner Pitfalls
CCEA Marking & Presentation Standards
• Formula Booklet: Most standard algebraic identities (indices laws, difference of two squares, factor rules) are not in the formula booklet. Memorise them completely.
• Show Full Working: In Unit AS 1, method marks require explicit algebraic steps. Writing down roots straight from a graphical or programmable calculator without intermediate lines of working loses all method marks.
• Degree of Accuracy: Unless exact values (surds/fractions) are required or specified otherwise, all non-exact numerical answers must be given to three significant figures (3 s.f.).
Top Examiner Traps to Avoid
1. Cancelling Terms Across Sums:
Never cancel individual terms across additions or subtractions (e.g., in \(\frac{x^2 + 5}{x}\), you cannot cancel \(x\)). You must factorise common factors first before cancelling.
2. Dividing Through by the Variable:
In equations like \(x^2 = 5x\), do not divide both sides by \(x\). This eliminates the valid root \(x = 0\). Instead, rearrange and factorise: \(x^2 - 5x = 0 \implies x(x - 5) = 0 \implies x = 0 \text{ or } x = 5\).
3. Confusing Conditions for Real Roots:
• "Two distinct real roots" \(\implies \Delta > 0\)
• "Real roots exist" \(\implies \Delta \geq 0\)
Using \(>\) when \(\geq\) is required loses accuracy marks.
4. Negative Sign Distribution in Surds:
When multiplying numerators and denominators by conjugates, be rigorous with brackets: \(-(a - \sqrt{b}) = -a + \sqrt{b}\).
Quick Chapter Summary Checklist
Before sitting your AS 1 exam, make sure you can confidently:
• Apply all 6 laws of indices and simplify surds.
• Rationalise denominators using algebraic conjugates.
• Complete the square and identify the turning point \((-p, q)\).
• Use the discriminant \(\Delta = b^2 - 4ac\) to classify roots and curve-line intersections.
• Solve quadratic inequalities and write the solution in correct disjoint or single-region notation.
• Solve simultaneous equations algebraically and interpret intersections geometrically.
• Apply the Factor and Remainder Theorems and perform polynomial long division.
• Sketch functions with all intercepts and asymptotes labelled, and apply all 6 function transformations accurately.