This note covers algebraic manipulation and factorising: collecting like terms, expanding brackets, factorising expressions, and (at Higher tier) simplifying algebraic fractions. These are the core skills for solving equations, rearranging formulae and working with graphs, so it pays to make them automatic.
Key ideas
Terms and expressions. An expression is made of terms joined by \(+\) and \(-\). Like terms have exactly the same letters raised to the same powers, so \(3x^2\) and \(5x^2\) are like terms but \(3x^2\) and \(3x\) are not. You can only add or subtract like terms.
Index laws in algebra. The same laws as for numbers apply: \(x^a\times x^b=x^{a+b}\), \(x^a\div x^b=x^{a-b}\), and \((x^a)^b=x^{ab}\). Multiply the number parts separately from the letters.
Expanding brackets. Multiply every term inside the bracket by the term outside. For two brackets, multiply each term in the first by each term in the second, then collect like terms. For example, \((x+3)(x+5)=x^2+5x+3x+15=x^2+8x+15\). A bracket squared means the bracket multiplied by itself, so \((x+4)^2=x^2+8x+16\), not \(x^2+16\). At Higher tier you may also be asked to expand three brackets; do two first, then multiply the result by the third.
Factorising is the reverse of expanding. Always look for these in order:
- Common factor. Take out the highest common factor of all terms, including letters. For example, \(12ab-18a^2b=6ab(2-3a)\).
- Difference of two squares. \(a^2-b^2=(a-b)(a+b)\), such as \(x^2-49=(x-7)(x+7)\). Look for a common factor first: \(3x^2-12=3(x-2)(x+2)\).
- Quadratics of the form \(x^2+bx+c\). Find two numbers that multiply to \(c\) and add to \(b\). For \(x^2+5x-14\), the numbers are \(7\) and \(-2\), giving \((x+7)(x-2)\).
- Quadratics of the form \(ax^2+bx+c\) (Higher tier). Find two numbers that multiply to \(ac\) and add to \(b\). Split the middle term using them, then factorise in pairs.
Always check a factorisation by expanding it again.
Algebraic fractions (Higher tier). To simplify an algebraic fraction, factorise the numerator and the denominator, then cancel common factors. You can only cancel factors, never terms. To add or subtract, find a common denominator and combine the numerators. To multiply, multiply tops and bottoms. To divide, turn the second fraction upside down and multiply.
Simplifying products and powers. To simplify a product such as \(3x^2y\times4xy^3\), multiply the numbers first (\(3\times4=12\)), then each letter by adding its indices (\(x^2\times x=x^3\) and \(y\times y^3=y^4\)). The result is \(12x^3y^4\). For a power of a product, such as \((2x^2)^3\), cube the number and multiply the index: \(8x^6\). For a quotient, divide the numbers and subtract the indices.
Choosing a method for factorising. Ask three questions in order. Is there a common factor? Is it a difference of two squares, which has two terms that are both perfect squares with a minus between them? Is it a quadratic with three terms? An expression with only two terms and no square structure, such as \(x^2+9\), usually cannot be factorised at this level.
Checking with numbers. If you are unsure whether two expressions are equivalent, substitute a simple value such as \(x=2\) into both. If the values differ, there is a mistake. If they match, you have good evidence, although it does not prove the result.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| Like terms | Terms with identical letter parts. These can be added or subtracted. |
| \((a+b)(c+d)=ac+ad+bc+bd\) | Expanding two brackets. |
| \((a+b)^2=a^2+2ab+b^2\) | Square of a sum. |
| \((a-b)^2=a^2-2ab+b^2\) | Square of a difference. |
| \(a^2-b^2=(a-b)(a+b)\) | Difference of two squares. |
| \(x^2+bx+c=(x+p)(x+q)\) | \(pq=c\) and \(p+q=b\). |
| \(ax^2+bx+c\) | Find two numbers with product \(ac\) and sum \(b\), split the \(x\) term, then factorise in pairs (Higher tier). |
| Algebraic fraction | A fraction with algebra in the numerator or denominator (Higher tier). |
Worked examples
Example 1 Expand and simplify \((2x-3)(x+5)-x(x-4)\).
Expand the first pair of brackets: \(2x\times x=2x^2\), \(2x\times5=10x\), \(-3\times x=-3x\), \(-3\times5=-15\). This gives \(2x^2+7x-15\).
Expand the second part: \(-x(x-4)=-x^2+4x\). Take care with the signs: the minus sign in front of \(x\) multiplies both terms.
Combine: \(2x^2+7x-15-x^2+4x\).
Collect like terms: \(x^2\) terms give \(2x^2-x^2=x^2\); \(x\) terms give \(7x+4x=11x\).
Answer: \(x^2+11x-15\).
Example 2 Factorise fully (a) (Higher) \(6x^2+x-15\) and (b) \(3x^2-48\).
(a) Here \(ac=6\times(-15)=-90\) and \(b=1\). We need two numbers with product \(-90\) and sum \(1\): these are \(10\) and \(-9\).
Split the middle term: \(6x^2+10x-9x-15\).
Factorise in pairs: \(2x(3x+5)-3(3x+5)\).
Take out the common bracket: \((3x+5)(2x-3)\).
Check: \((3x+5)(2x-3)=6x^2-9x+10x-15=6x^2+x-15\). Correct.
(b) Take out the common factor 3 first: \(3(x^2-16)\). Then use the difference of two squares: \(3(x-4)(x+4)\).
Answer: (a) \((3x+5)(2x-3)\); (b) \(3(x-4)(x+4)\).
Example 3 (Higher) (a) Simplify \(\dfrac{x^2-9}{x^2+x-6}\). (b) Write \(\dfrac{2}{x+1}+\dfrac{3}{x-2}\) as a single fraction.
(a) Factorise the numerator using the difference of two squares: \(x^2-9=(x-3)(x+3)\).
Factorise the denominator: two numbers with product \(-6\) and sum \(1\) are \(3\) and \(-2\), so \(x^2+x-6=(x+3)(x-2)\).
Cancel the common factor \((x+3)\): \(\dfrac{(x-3)(x+3)}{(x+3)(x-2)}=\dfrac{x-3}{x-2}\).
(b) The common denominator is \((x+1)(x-2)\).
Write each fraction over it: \(\dfrac{2(x-2)}{(x+1)(x-2)}+\dfrac{3(x+1)}{(x+1)(x-2)}\).
Add the numerators: \(2x-4+3x+3=5x-1\).
Answer: (a) \(\dfrac{x-3}{x-2}\); (b) \(\dfrac{5x-1}{(x+1)(x-2)}\).
Common errors
- Writing \((x+4)^2=x^2+16\). The middle term \(8x\) is missing, so expand as two brackets.
- Sign mistakes when a negative sits outside a bracket, such as \(-x(x-4)=-x^2-4x\). The correct result is \(-x^2+4x\).
- Not taking out a common factor first, so a quadratic such as \(2x^2+10x+12\) is factorised as a harder problem than \(2(x+2)(x+3)\).
- Factorising only partly. For example, \(4x^2-16\) as \((2x-4)(2x+4)\) still has common factors; the full answer is \(4(x-2)(x+2)\).
- Cancelling terms instead of factors, such as cancelling the \(x\) in \(\dfrac{x+3}{x}\).
Exam tip: when a question says "factorise fully", look for a common factor before anything else, and expand your answer to check it. If an expression has two terms, try the difference of two squares.
Check yourself
- Expand and simplify \((x+4)(x-7)\).
Answer
\(x^2-7x+4x-28=x^2-3x-28\). - (Higher) Factorise \(2x^2+7x+3\).
Answer
\(ac=6\) and \(b=7\), so use 6 and 1: \(2x^2+6x+x+3=2x(x+3)+1(x+3)=(2x+1)(x+3)\). - Factorise fully \(12ab-18a^2b\).
Answer
The highest common factor is \(6ab\), so the expression is \(6ab(2-3a)\). - (Higher) Simplify \(\dfrac{2x^2+6x}{x^2-9}\).
Answer
Numerator: \(2x(x+3)\). Denominator: \((x-3)(x+3)\). Cancel \((x+3)\) to get \(\dfrac{2x}{x-3}\).