This Year 10 algebraic notation, expanding and factorising revision notes is written for pupils aged 14-15 and follows the National Curriculum. In short, it shows you how to collect terms, expand brackets and factorise including x^2 + bx + c and the difference of two squares. Before you start, it helps to revise surds and exact values and units, estimation, rounding and bounds, because this lesson builds on them.
Year 10 algebraic notation, expanding and factorising revision notes: what you will learn
Algebra is the language of GCSE maths. This chapter covers writing algebra correctly, substituting, collecting like terms, expanding brackets and factorising. It leads straight into equations, quadratics and proof. Factorising \(ax^2+bx+c\) with \(a\neq1\) and algebraic fractions are Higher tier only.
Starter activity
Imagine you buy 3 packets of crisps at \(c\) pence each and 2 bottles of water at \(w\) pence each. Write an expression for the total cost. Now your friend buys 5 packets of crisps and 4 bottles. Write their cost and add the two bills. Which terms could you combine? Which could you not combine, and why? Try again with \(c=80\) and \(w=65\) to check that your simplified expression gives the same total.
Notation and vocabulary
Definition: algebraic notation and vocabulary
- \(3a\) means \(3\times a\); \(ab\) means \(a\times b\); \(a^2\) means \(a\times a\); \(\dfrac{a}{b}\) means \(a\div b\).
- A term is a number, letter or product, such as \(5x\) or \(-3y^2\). An expression has no equals sign, an equation has one, a formula links variables, and an identity is true for all values (written \(\equiv\)).
- Write the number first and letters in alphabetical order: \(5ab\), not \(b5a\).
Substitution
Rule: substitution
Replace each letter by its value, using brackets for negative numbers. If \(x=-2\) then \(3x^2-4x+1=3(-2)^2-4(-2)+1=12+8+1=21\). Remember that \((-2)^2=4\), not \(-4\).
Simplifying
Definition: like terms
Like terms have the same letters with the same powers, such as \(3a\) and \(-5a\), or \(2x^2\) and \(7x^2\). Only like terms can be collected.
Key fact: collecting like terms
Collect like terms (same letters with the same powers). \(3a+2b-a+5b=2a+7b\). For products multiply the numbers then the letters: \(4x\times3y=12xy\). For quotients cancel common factors: \(\dfrac{6x^2y}{2xy}=3x\).
Expanding brackets
Rule: expanding brackets
Multiply every term inside the bracket by the term outside: \(3(2x-5)=6x-15\). With two brackets multiply each term in the first by each term in the second (FOIL or a grid) and collect like terms.
\((x+4)(x+5)=x^2+5x+4x+20=x^2+9x+20\)
\((x-3)(x+7)=x^2+7x-3x-21=x^2+4x-21\)
\((2x+1)(x-6)=2x^2-12x+x-6=2x^2-11x-6\)
Squaring a bracket means writing it twice: \((x-5)^2=(x-5)(x-5)=x^2-10x+25\). It is not \(x^2+25\).
Method: expanding double brackets
- Multiply every term in the first bracket by every term in the second (FOIL or a grid).
- Write out all four products.
- Collect the like terms.
Factorising
Definition: factorising
Factorising is expanding in reverse. Always look for a common factor first and take out the highest one: \(12x+18=6(2x+3)\) and \(5x^2-15x=5x(x-3)\).
Factorising \(x^2+bx+c\)
Method: factorising x squared plus bx plus c
Find two numbers that multiply to \(c\) and add to \(b\).
- \(x^2+7x+12\): numbers 3 and 4, so \((x+3)(x+4)\).
- \(x^2-5x-14\): numbers \(-7\) and 2 (product \(-14\), sum \(-5\)), so \((x-7)(x+2)\).
Check by expanding: the middle term and the last term should match.
Difference of two squares
Rule: the difference of two squares
\(a^2-b^2=(a-b)(a+b)\). So \(x^2-49=(x-7)(x+7)\) and \(9y^2-16=(3y-4)(3y+4)\). There must be a minus between two perfect squares.
Factorising \(ax^2+bx+c\) (Higher tier)
Method: factorising when a is not 1 (Higher tier)
Multiply \(a\times c\), find two numbers that multiply to \(ac\) and add to \(b\), split the middle term and factorise in pairs. For \(2x^2+7x+3\): \(ac=6\), numbers 1 and 6. \(2x^2+x+6x+3=x(2x+1)+3(2x+1)=(2x+1)(x+3)\).
Algebraic fractions (Higher tier)
Rule: simplifying algebraic fractions (Higher tier)
Factorise the top and bottom completely, then cancel common factors (never common terms). \(\dfrac{x^2-9}{x^2+5x+6}=\dfrac{(x-3)(x+3)}{(x+2)(x+3)}=\dfrac{x-3}{x+2}\).
Worked example in exam style
Worked example 1
Show that \((n+3)^2-9=n(n+6)\) [2 marks].
Expand: \(n^2+6n+9-9=n^2+6n=n(n+6)\). Each line is shown, ending with the form asked for.
More worked examples
Worked example 2
Expand and simplify \((3x-2)(x+4)-5x\) [3 marks].
\(3x^2+12x-2x-8-5x=3x^2+5x-8\). Notice that the \(-5x\) is collected only after the brackets are expanded.
Worked example 3
Factorise fully \(8x^2y-12xy^2\) [2 marks].
The highest common factor of 8 and 12 is 4, and both terms contain \(x\) and \(y\). \(4xy(2x-3y)\). Check: \(4xy\times2x=8x^2y\) and \(4xy\times3y=12xy^2\).
Worked example 4
Factorise \(x^2-6x+9\) [2 marks].
Numbers with product 9 and sum \(-6\): \(-3\) and \(-3\). \((x-3)^2\). This is a perfect square, so the two brackets are the same.
Worked example 5
A square has side \((2x+1)\) cm. Show that its area is \(4x^2+4x+1\) [2 marks].
\((2x+1)^2=4x^2+2x+2x+1=4x^2+4x+1\).
Worked example 6 (Higher tier)
Simplify \(\dfrac{3x^2+x-2}{3x^2-5x+2}\) [4 marks].
Top: \(ac=-6\), numbers 3 and \(-2\): \(3x^2+3x-2x-2=3x(x+1)-2(x+1)=(3x-2)(x+1)\). Bottom: \(ac=6\), numbers \(-3\) and \(-2\): \(3x^2-3x-2x+2=3x(x-1)-2(x-1)=(3x-2)(x-1)\). Cancel \((3x-2)\): \(\dfrac{x+1}{x-1}\).
Common mistakes
Watch out
- Writing \((x+3)^2=x^2+9\): the middle term \(6x\) is missing.
- Sign slips when expanding a negative: \(-2(x-1)=-2x+2\), not \(-2x-2\).
- Not taking out the highest common factor: \(6x^2+9x\) is \(3x(2x+3)\), not \(x(6x+9)\).
- Cancelling terms instead of factors in a fraction, such as cancelling the \(x\) in \(\frac{x+3}{x}\).
Exam technique
Exam technique
- Write each expansion on its own line and circle the like terms before collecting them.
- Factorise "fully" means no further common factor is left: check each bracket.
- Substitution questions often use negative values: write the brackets in.
- If a question says "show that", every line must be visible and the final line must match the target exactly.
Building and checking expressions
Algebra works like a language with its own grammar. A formula or expression is built from letters that stand for numbers, so every rule of arithmetic carries over. The three habits that make algebra reliable are writing each step on its own line, using brackets when you substitute negatives, and checking your result with a simple number. For example, to check that \((x+3)(x+4)=x^2+7x+12\), try \(x=2\): the left side is \(5\times6=30\) and the right side is \(4+14+12=30\). A match does not prove the identity, but a mismatch proves that you made an error.
Worked example 7: substituting into a formula
Use \(v=u+at\) to find \(v\) when \(u=-4\), \(a=3\) and \(t=5\) (non-calculator).
Replace each letter with its value, using brackets for the negative: \(v=(-4)+3\times5\). Multiply first: \(3\times5=15\). Then add: \(-4+15=11\). So \(v=11\). If you wrote \(-4+3\times5\) as \((-4+3)\times5\) you would get \(-5\), which shows why the order of operations (BIDMAS) matters.
Worked example 8: expanding with a negative in front
Expand and simplify \(3(x+2)-2(x-5)\) [3 marks].
Expand the first bracket: \(3x+6\). Expand the second bracket: \(-2\times x=-2x\) and \(-2\times(-5)=+10\). Now collect like terms: \(3x+6-2x+10=x+16\). The common slip is to write \(-2x-10\); the minus sign in front of the bracket changes the sign of every term inside it.
Worked example 9: the square of a bracket
Expand \((3x-2)^2\) and then check with \(x=1\).
Write it as two brackets: \((3x-2)(3x-2)\). Multiply: \(9x^2-6x-6x+4=9x^2-12x+4\). Check with \(x=1\): left side \((3-2)^2=1\); right side \(9-12+4=1\). Good. The wrong answer \(9x^2+4\) would give \(13\), which does not match.
Worked example 10: factorising in two stages
Factorise fully \(2x^2-18\) [3 marks].
First take out the common factor 2: \(2(x^2-9)\). Now \(x^2-9\) is a difference of two squares, \(x^2-3^2\), so it factorises as \((x-3)(x+3)\). The full answer is \(2(x-3)(x+3)\). Many students stop after the first stage, but a question that says “fully” expects both.
Worked example 11: forming and simplifying in context
A cinema charges £\(a\) for an adult and £\(c\) for a child. A family buys 2 adult and 3 child tickets and a group buys 5 adult and 4 child tickets. Write the total cost of both groups and simplify.
Family: \(2a+3c\). Group: \(5a+4c\). Total: \(2a+3c+5a+4c=7a+7c=7(a+c)\). The factorised form tells us that the combined bill equals the cost of 7 pairs of one adult and one child ticket.
Note: expanding a longer bracket
Whether you use the grid method, the “smile and eyebrows” method or FOIL, all the products must appear before you collect like terms. A grid is the safest method for longer brackets such as \((x+2)(x^2-3x+4)\), because each term in the first bracket multiplies each term in the second: \(x^3-3x^2+4x+2x^2-6x+8=x^3-x^2-2x+8\). Count your products: a 2-term bracket times a 3-term bracket must give \(2\times3=6\) products.
Choosing the right tool
Many GCSE questions in this topic use the command words simplify, expand, factorise and show that, and each one asks for a different action. To simplify you collect like terms and the expression gets shorter. To expand you multiply out brackets and the expression usually gets longer. To factorise you do the reverse of expanding and put brackets back in. To show that you start from one side and reach the other, writing every line, without using the answer you are trying to prove. If you are unsure which action is needed, look at the shape of the expression: brackets with nothing outside need expanding, a long sum with a common factor needs factorising, and a sum of several terms with repeated letters needs simplifying. Practising the four actions in mixed order, rather than one at a time, is the best way to prepare for a real exam paper.
Watch out: signs and squares
Two sign errors cost marks more than any others. First, a minus sign in front of a bracket must be applied to every term, so \(-(2x-3)=-2x+3\). Second, \((-x)^2=x^2\) but \(-x^2\) means \(-(x^2)\), so with \(x=4\) the first is \(16\) and the second is \(-16\). Write brackets round negative numbers on your calculator too.
Did you know?
The word algebra comes from the Arabic al-jabr, from a book written around the year 820 by the mathematician Muhammad ibn Musa al-Khwarizmi in Baghdad. “Al-jabr” means “restoring” or “completing”, which describes the step of moving a subtracted term to the other side. Our word algorithm also comes from his name. The use of letters for unknown numbers came much later: the French mathematician François Viète used letters in the 1590s, and René Descartes introduced the habit of using \(x\), \(y\) and \(z\) for unknowns and \(a\), \(b\) and \(c\) for known values. Every time you write \(3x+4\) you are using ideas that took more than 800 years to develop.
Otto’s tip
Always check a factorisation by expanding it. If it does not give back the original, change the signs of your two numbers and try again.
How to use this Year 10 algebraic notation, expanding and factorising revision notes
First, read the key points so that you know where the Year 10 algebraic notation, expanding and factorising revision notes is going. Then work through each definition and worked example, and cover the solution before you read it. After that, practise with the algebraic notation, expanding and factorising worksheet and check every question with the worksheet answers. When you feel ready, test yourself with the multiple choice questions or the quick quiz. Finally, sit the algebraic notation, expanding and factorising test and mark it with the test answers.
Also, the starters are a good warm-up, so you can keep the key facts fresh. Moreover, all the pages of this chapter work together, so you can move between them at any time. For example, many pupils read the lesson on Monday, practise on Wednesday and test themselves on Friday. In short, this Year 10 algebraic notation, expanding and factorising revision notes gives you everything you need for the chapter.
To sum up, regular practice works better than one long session. For instance, ten minutes a day is often enough. Besides, mistakes are useful, because they show what to revise next. Then, once a method feels easy, try a harder question. As a result, you build confidence step by step.
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