This note covers expanding brackets, factorising expressions in several ways, and working with algebraic fractions. These are the core manipulation skills behind solving equations, sketching graphs and simplifying formulae. Algebraic fractions are mainly an Extended topic.
Key ideas
Expanding. To expand, multiply every term in one bracket by every term in the other. For a single bracket, \(3x(2x-5)=6x^2-15x\). For two brackets, each of the four products is needed, then like terms are collected. A square such as \((x-4)^2\) means two copies, so it needs all four products, not just \(x^2+16\). Extended work may involve three brackets: expand two, simplify, then multiply by the third.
Factorising. Factorising is the reverse of expanding. Always check the following in order.
- Common factor. Take out the highest common factor of all the terms first: \(6x^2y-9xy=3xy(2x-3)\).
- Difference of two squares. \(a^2-b^2=(a-b)(a+b)\). This also works with numbers and coefficients, such as \(49a^2-16b^2=(7a-4b)(7a+4b)\).
- Quadratic trinomials. For \(x^2+bx+c\), find two numbers that multiply to \(c\) and add to \(b\). For \(ax^2+bx+c\), find two numbers that multiply to \(ac\) and add to \(b\), split the middle term, then factorise in pairs.
- Grouping. With four terms, pair them up so that each pair has a common factor, and look for a common bracket.
A trinomial such as \(x^2+10x+25\) is a perfect square: it factorises as \((x+5)^2\). Spotting these saves time. If the quadratic has a negative leading coefficient or a common factor, take that out first so that you are factorising a simpler expression.
Always multiply back out to check a factorisation. Expanding and factorising are opposite processes, so checking one with the other is a reliable habit. A good plan is to decide which method fits by counting terms: two terms suggests a common factor or a difference of two squares, three terms suggests a trinomial, and four terms suggests grouping.
Algebraic fractions (Extended). They follow the same rules as numerical fractions.
- To simplify, factorise the numerator and the denominator, then cancel common factors. You can only cancel factors (things multiplied), never terms (things added).
- To multiply, multiply tops and bottoms after factorising and cancelling. To divide, multiply by the reciprocal of the second fraction.
- To add or subtract, find a common denominator, usually the product of the different denominators, then combine the numerators. Take care with a minus sign in front of a bracket.
Algebraic fractions are not defined when the denominator is zero, so any value of \(x\) that makes a denominator zero is not allowed.
For example, to divide \(\dfrac{x+1}{x-2}\div\dfrac{x^2-1}{x-2}\), flip the second fraction and multiply: \(\dfrac{x+1}{x-2}\times\dfrac{x-2}{(x-1)(x+1)}\). The factors \((x+1)\) and \((x-2)\) cancel, leaving \(\dfrac{1}{x-1}\). Factorising first almost always makes the work shorter and keeps the numbers small.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| Expand | Remove brackets by multiplying out. |
| Factorise | Write an expression as a product of simpler expressions. |
| \((a+b)^2=a^2+2ab+b^2\) | Square of a sum. The middle term is twice the product. |
| \((a-b)^2=a^2-2ab+b^2\) | Square of a difference. |
| \(a^2-b^2=(a-b)(a+b)\) | Difference of two squares. |
| \(x^2+(p+q)x+pq=(x+p)(x+q)\) | Factorising a quadratic with leading coefficient 1. |
| \(\dfrac{a}{b}\pm\dfrac{c}{d}=\dfrac{ad\pm bc}{bd}\) | Adding or subtracting fractions with different denominators. |
| \(\dfrac{a}{b}\div\dfrac{c}{d}=\dfrac{a}{b}\times\dfrac{d}{c}\) | Dividing by a fraction. |
Worked examples
Example 1 The area of a rectangular solar panel is \((3x-2)(2x+5)\) cm². (a) Expand and simplify this expression. (b) (Extended) Expand and simplify \((x+2)(x-1)(x+3)\).
Multiply every term in the first bracket by every term in the second:
\(3x\times2x=6x^2\), \(3x\times5=15x\), \(-2\times2x=-4x\), \(-2\times5=-10\).
Collect like terms: \(15x-4x=11x\), so the product is \(6x^2+11x-10\).
(b) Expand the first two brackets: \((x+2)(x-1)=x^2-x+2x-2=x^2+x-2\).
Now multiply by \((x+3)\): \(x^2\times x=x^3\), \(x^2\times3=3x^2\), \(x\times x=x^2\), \(x\times3=3x\), \(-2\times x=-2x\), \(-2\times3=-6\).
Collect like terms: \(3x^2+x^2=4x^2\) and \(3x-2x=x\). Check by substituting \(x=1\): the original has a factor \((x-1)=0\), and \(1+4+1-6=0\). Correct.
Answer: (a) \(6x^2+11x-10\); (b) \(x^3+4x^2+x-6\)
Example 2 Factorise fully: (a) \(6x^2-7x-3\); (b) \(2xy+6x-y-3\); (c) \(49a^2-16b^2\).
(a) Multiply the first and last coefficients: \(6\times(-3)=-18\). We need two numbers that multiply to \(-18\) and add to \(-7\): these are \(-9\) and \(2\).
Split the middle term: \(6x^2-9x+2x-3\). Factorise in pairs: \(3x(2x-3)+1(2x-3)\).
Take out the common bracket: \((2x-3)(3x+1)\).
Check: \(6x^2+2x-9x-3=6x^2-7x-3\). Correct.
(b) Group in pairs: \(2x(y+3)-1(y+3)\). The common bracket is \((y+3)\), so the result is \((y+3)(2x-1)\).
(c) Both terms are perfect squares: \((7a)^2-(4b)^2=(7a-4b)(7a+4b)\).
Answer: (a) \((2x-3)(3x+1)\); (b) \((y+3)(2x-1)\); (c) \((7a-4b)(7a+4b)\)
Example 3 (Extended) (a) Simplify \(\dfrac{x^2-9}{x^2+5x+6}\). (b) Write \(\dfrac{3}{x+1}-\dfrac{2}{x-2}\) as a single fraction.
(a) Factorise the top as a difference of two squares: \((x-3)(x+3)\). Factorise the bottom: two numbers with product 6 and sum 5 are 2 and 3, so \((x+2)(x+3)\).
\[\frac{(x-3)(x+3)}{(x+2)(x+3)}\]
Cancel the common factor \((x+3)\): \(\dfrac{x-3}{x+2}\).
(b) A common denominator is \((x+1)(x-2)\). Write each fraction over it:
\[\frac{3(x-2)}{(x+1)(x-2)}-\frac{2(x+1)}{(x+1)(x-2)}=\frac{3(x-2)-2(x+1)}{(x+1)(x-2)}\]
Expand the top, taking care with the minus sign: \(3x-6-2x-2=x-8\).
Answer: (a) \(\dfrac{x-3}{x+2}\); (b) \(\dfrac{x-8}{(x+1)(x-2)}\)
Common errors
- Expanding a square as \((x+3)^2=x^2+9\). The middle term \(6x\) is missing: \((x+3)^2=x^2+6x+9\).
- Sign errors when a negative multiplies a bracket, such as \(-2(x+1)=-2x-2\), not \(-2x+2\).
- Cancelling terms instead of factors, such as writing \(\dfrac{x+3}{x}=3\) or cancelling the \(x^2\) in \(\dfrac{x^2-9}{x^2+5x+6}\). Only brackets that multiply can be cancelled.
- Stopping before an expression is fully factorised, for example leaving \(2x^2-8\) as \(2(x^2-4)\) when it is \(2(x-2)(x+2)\).
- Adding fractions by adding the denominators. Common denominators are needed first.
Exam tip: After factorising, multiply the brackets back out in your head or in the margin. It takes ten seconds and catches almost every sign slip.
Check yourself
- Expand and simplify \((x-4)^2\).
Answer
\((x-4)(x-4)=x^2-4x-4x+16=x^2-8x+16\). - Factorise fully \(3x^2-12\).
Answer
Take out 3: \(3(x^2-4)\). Then difference of two squares: \(3(x-2)(x+2)\). - Factorise \(x^2+x-12\).
Answer
Two numbers with product \(-12\) and sum 1 are 4 and \(-3\). So \((x+4)(x-3)\). - (Extended) Simplify \(\dfrac{x^2-4x}{x^2-16}\), and write \(\dfrac{1}{x}+\dfrac{1}{x+2}\) as one fraction.
Answer
First: \(\dfrac{x(x-4)}{(x-4)(x+4)}=\dfrac{x}{x+4}\). Second: \(\dfrac{(x+2)+x}{x(x+2)}=\dfrac{2x+2}{x(x+2)}\).