Welcome to Algebraic Manipulation!
Welcome to one of the most powerful and essential chapters in Unit 1: Pure Mathematics! If you think of mathematics as a language, algebra is its grammar. Mastering algebraic manipulation gives you the tools to solve complex equations, sketch curves, and unlock advanced calculus topics later on. Don't worry if algebra has felt intimidating before; we will break down every single technique into clear, manageable steps with plenty of examples!
1. Expanding Brackets (Products of Binomials)
Expanding means multiplying out brackets to remove them. In CCEA GCSE Further Mathematics, you will often need to expand products of three brackets or expressions with higher powers.
Expanding Two Brackets
To multiply two linear terms like \((ax + b)(cx + d)\), ensure every term in the first bracket multiplies every term in the second bracket. Many people use the FOIL memory aid:
• First terms
• Outside terms
• Inside terms
• Last terms
Example: Expand and simplify \((2x - 3)(3x + 4)\)
• First: \(2x \times 3x = 6x^2\)
• Outside: \(2x \times 4 = 8x\)
• Inside: \(-3 \times 3x = -9x\)
• Last: \(-3 \times 4 = -12\)
Combine like terms: \(6x^2 + 8x - 9x - 12 = 6x^2 - x - 12\)
Expanding Three Brackets
When you have three brackets such as \((x + 1)(x - 2)(2x + 3)\), do it in two clear stages:
Step 1: Choose any two brackets and multiply them together first.
Step 2: Multiply the resulting quadratic expression by the third bracket.
Step 3: Collect all like terms.
Example: Expand and simplify \((x + 2)(x - 3)(2x - 1)\)
• First, expand \((x + 2)(x - 3) = x^2 - 3x + 2x - 6 = x^2 - x - 6\)
• Now multiply by the third bracket: \((x^2 - x - 6)(2x - 1)\)
• Multiply each term in the quadratic by \(2x\): \(2x(x^2 - x - 6) = 2x^3 - 2x^2 - 12x\)
• Multiply each term in the quadratic by \(-1\): \(-1(x^2 - x - 6) = -x^2 + x + 6\)
• Add them together: \((2x^3 - 2x^2 - 12x) + (-x^2 + x + 6) = 2x^3 - 3x^2 - 11x + 6\)
Key Takeaway: Take your time with negative signs! Expanding three brackets is just doing standard expansion twice.
2. Advanced Factorisation
Factorising is the exact opposite of expanding: it means writing an expression as a product of its factors (putting brackets back in). Always look for common factors first!
A. Highest Common Factor (HCF) and Factorising by Grouping
When an expression has four terms, you can often factorise by grouping them in pairs.
Example: Factorise \(6x^2 - 4ax + 9bx - 6ab\)
• Group into two pairs: \((6x^2 - 4ax) + (9bx - 6ab)\)
• Factor out the HCF from each pair: \(2x(3x - 2a) + 3b(3x - 2a)\)
• Notice that \((3x - 2a)\) is a shared common bracket! Factor it out: \((3x - 2a)(2x + 3b)\)
B. Difference of Two Squares (DOTS)
Remember the rule: \(a^2 - b^2 \equiv (a - b)(a + b)\)
Example: Factorise \(16x^2 - 81y^2\)
• Identify the square roots: \(\sqrt{16x^2} = 4x\) and \(\sqrt{81y^2} = 9y\)
• Answer: \((4x - 9y)(4x + 9y)\)
C. Quadratic Expressions of the form \(ax^2 + bx + c\)
When the coefficient \(a \neq 1\), use the "ac method":
Step 1: Multiply \(a\) and \(c\) to get the product \(ac\).
Step 2: Find two numbers that multiply to give \(ac\) and add together to give \(b\).
Step 3: Split the middle term \(bx\) into two terms using these numbers.
Step 4: Factorise by grouping.
Example: Factorise \(3x^2 + 11x - 4\)
• Product: \(a \times c = 3 \times (-4) = -12\). Sum: \(b = 11\).
• Two numbers that multiply to \(-12\) and add to \(11\) are \(12\) and \(-1\).
• Split the middle term: \(3x^2 + 12x - 1x - 4\)
• Factorise by grouping: \(3x(x + 4) - 1(x + 4) = (3x - 1)(x + 4)\)
Key Takeaway: Always check your factorisation by multiplying the brackets back out in your head to ensure you get the original expression.
3. Completing the Square
Completing the square rewrites a quadratic into the vertex form \(a(x + p)^2 + q\). This is useful for finding maximum/minimum turning points and solving quadratic equations.
Case 1: When \(a = 1\) (Monic Quadratic \(x^2 + bx + c\))
The standard formula is: \(x^2 + bx + c = \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c\)
Example: Express \(x^2 + 6x - 5\) in the form \((x + p)^2 + q\)
• Halve the \(x\)-coefficient: \(\frac{6}{2} = 3\)
• Write: \((x + 3)^2 - (3)^2 - 5\)
• Simplify the numbers: \((x + 3)^2 - 9 - 5 = (x + 3)^2 - 14\)
Case 2: When \(a \neq 1\) (Non-monic Quadratic \(ax^2 + bx + c\))
Step 1: Factor out the coefficient \(a\) from the terms containing \(x^2\) and \(x\).
Step 2: Complete the square inside the square bracket.
Step 3: Multiply back out by \(a\) and collect numerical constants.
Example: Express \(2x^2 - 12x + 11\) in the form \(a(x + p)^2 + q\)
• Factor out \(2\) from the first two terms: \(2[x^2 - 6x] + 11\)
• Complete the square inside the bracket: \(2\left[(x - 3)^2 - 3^2\right] + 11 = 2\left[(x - 3)^2 - 9\right] + 11\)
• Multiply by \(2\): \(2(x - 3)^2 - 18 + 11\)
• Simplify: \(2(x - 3)^2 - 7\)
Did you know? From \(2(x - 3)^2 - 7\), we can instantly see that the minimum value of this curve is \(-7\), and it occurs when \(x = 3\)!
Key Takeaway: Remember to subtract \(\left(\frac{b}{2}\right)^2\) inside the bracket and multiply it by \(a\) when expanding out!
4. Algebraic Fractions
Algebraic fractions follow the exact same arithmetic rules as numerical fractions: find a common denominator for addition/subtraction, and cancel common factors when simplifying, multiplying, or dividing.
A. Simplifying Algebraic Fractions
Crucial Rule: Never cancel individual terms across a plus or minus sign! You must factorise completely first, then cancel common factors.
Example: Simplify \(\frac{2x^2 - 8}{x^2 + 3x - 10}\)
• Factorise numerator: \(2(x^2 - 4) = 2(x - 2)(x + 2)\)
• Factorise denominator: \((x + 5)(x - 2)\)
• Cancel the common factor \((x - 2)\): \(\frac{2(x + 2)}{x + 5}\)
B. Multiplication and Division
• Multiplication: Factorise numerators and denominators, multiply straight across, and cancel common factors.
• Division: Invert the second fraction (turn it upside down) and change the sign to multiplication ("Keep, Change, Flip").
Example: Simplify \(\frac{x^2 - 9}{2x + 4} \div \frac{x - 3}{x + 2}\)
• Flip the second fraction and multiply: \(\frac{x^2 - 9}{2x + 4} \times \frac{x + 2}{x - 3}\)
• Factorise all parts: \(\frac{(x - 3)(x + 3)}{2(x + 2)} \times \frac{x + 2}{x - 3}\)
• Cancel common terms \((x - 3)\) and \((x + 2)\): \(\frac{x + 3}{2}\)
C. Addition and Subtraction
To add or subtract fractions, find the Lowest Common Denominator (LCD).
Example: Express as a single fraction in its simplest form: \(\frac{3}{x + 2} - \frac{2}{x - 1}\)
• Common denominator is \((x + 2)(x - 1)\)
• Rewrite each fraction: \(\frac{3(x - 1)}{(x + 2)(x - 1)} - \frac{2(x + 2)}{(x + 2)(x - 1)}\)
• Combine over a single denominator: \(\frac{3(x - 1) - 2(x + 2)}{(x + 2)(x - 1)}\)
• Expand the numerator: \(\frac{3x - 3 - 2x - 4}{(x + 2)(x - 1)} = \frac{x - 7}{(x + 2)(x - 1)}\)
Common Pitfall to Avoid: Watch the minus sign when subtracting! In \(-2(x + 2)\), remember that \(-2 \times +2 = -4\), not \(+4\).
5. Changing the Subject of a Formula
Rearranging a formula means isolating a specific variable on one side of the equals sign. In Further Maths, the target variable will often appear more than once or be trapped inside brackets, fractions, or square roots.
Step-by-Step Strategy for Difficult Rearrangements:
1. Clear fractions: Multiply every term by the denominator.
2. Clear brackets: Expand all brackets containing the desired subject.
3. Group terms: Collect all terms containing the subject on one side of the equation, and move all other terms to the other side.
4. Factorise: Take out the subject as a common factor.
5. Divide: Divide by the remaining bracket to isolate the subject.
Example: Make \(x\) the subject of the formula \(y = \frac{2x + 5}{3 - x}\)
• Step 1 (Clear fractions): \(y(3 - x) = 2x + 5\)
• Step 2 (Expand brackets): \(3y - xy = 2x + 5\)
• Step 3 (Collect \(x\) terms on one side): \(3y - 5 = 2x + xy\)
• Step 4 (Factorise \(x\)): \(3y - 5 = x(2 + y)\)
• Step 5 (Divide): \(x = \frac{3y - 5}{2 + y}\)
Key Takeaway: When the subject appears twice, factorisation is almost always the key step to getting it on its own!
6. Algebraic Proof and Identities
An identity is an equation that is true for all values of the variable. We use the identity symbol \(\equiv\) instead of a standard equals sign \(=\).
Understanding Number Patterns in Proofs
To prove properties algebraically, we write general expressions for different types of numbers:
• Any integer: \(n\)
• Consecutive integers: \(n, n + 1, n + 2, \dots\)
• An even number: \(2n\) (since it is a multiple of \(2\))
• An odd number: \(2n + 1\) or \(2n - 1\)
• Consecutive even numbers: \(2n, 2n + 2\)
• Consecutive odd numbers: \(2n + 1, 2n + 3\)
• A multiple of \(k\): \(k \times m\) or \(k(m)\)
How to Structure a Proof
• State your definitions clearly (e.g., "Let the two consecutive integers be \(n\) and \(n + 1\)").
• Perform the algebraic expansion and simplification step by step.
• Factorise or rewrite the final expression to show the required property.
• Write a concluding sentence stating that the proof is complete.
Example: Prove that the difference between the squares of any two consecutive odd integers is always a multiple of \(8\).
• Let the consecutive odd integers be \(2n + 1\) and \(2n + 3\), where \(n\) is an integer.
• Set up the difference of their squares: \((2n + 3)^2 - (2n + 1)^2\)
• Expand both brackets: \((4n^2 + 12n + 9) - (4n^2 + 4n + 1)\)
• Simplify carefully: \(4n^2 + 12n + 9 - 4n^2 - 4n - 1 = 8n + 8\)
• Factor out \(8\): \(8(n + 1)\)
• Conclusion: Since \(8(n + 1)\) is divisible by \(8\) for any integer \(n\), the difference between the squares of any two consecutive odd integers is always a multiple of \(8\).
Quick Review: Chapter Checklist
Before moving on to the next chapter, check that you can:
• Expand three sets of brackets accurately.
• Factorise quadratic expressions \(ax^2 + bx + c\) and expressions with four terms by grouping.
• Complete the square for both monic (\(a = 1\)) and non-monic (\(a \neq 1\)) quadratics.
• Simplify, multiply, divide, add, and subtract complex algebraic fractions.
• Rearrange formulas where the subject appears multiple times or in denominators.
• Construct rigorous algebraic proofs using clear notation and algebraic definitions.