This Year 8 algebra lesson is written for pupils aged 12-13 and follows the National Curriculum. In short, it shows you how to expand and factorise expressions and use index laws with letters. Before you start, it helps to revise fractions, decimals and percentages and powers, roots and standard form, because this lesson builds on them.
Year 8 algebra lesson: what you will learn
This lesson covers the language of algebra, collecting like terms, the index laws for letters, expanding brackets and factorising. Read each box in order and try every worked example before you read the solution.
Starter activity
Think of a number, double it, add 6, then halve your answer. Try it with 4, 10 and 25. What do you notice about each answer compared with the starting number? Use a letter \(n\) for the number to explain why this always works.
The language of algebra
Definition: term and expression
A term is a single part of an expression, such as \(3x\) or \(-5\). An expression is a collection of terms with no equals sign, for example \(4x+3y-7\). The sign in front of a term belongs to that term.
Definition: like terms and coefficient
Like terms have exactly the same letters raised to the same powers, so \(3x\) and \(5x\) are like terms but \(3x\) and \(3x^2\) are not. The number in front of the letters is the coefficient: in \(3x\) the coefficient is 3.
Simplifying by collecting like terms
Rule: collecting like terms
Add or subtract the coefficients of like terms and keep the letter part unchanged. Unlike terms cannot be combined.
Worked example 1: collecting like terms
Simplify \(4x+3y-x+2y\).
Group the like terms: \((4x-x)+(3y+2y)\). Then \(4x-x=3x\) and \(3y+2y=5y\), so the answer is \(3x+5y\).
Multiplying and dividing terms
Key fact: multiplying and dividing terms
- Multiply the numbers, then the letters: \(3a\times4b=12ab\).
- Powers of the same letter: \(x^2\times x^3=x^5\) (add the indices).
- Division: \(12x^5\div3x^2=4x^3\) (divide the numbers, subtract the indices).
- Power of a power: \((2a^3)^2=2^2\times a^6=4a^6\). The number is squared too.
Worked example 2: index laws with letters
Simplify \(\dfrac{15a^4b}{5a^2}\) and \((3x^2)^3\).
Divide the numbers: \(15\div5=3\). Subtract the indices of \(a\): \(a^4\div a^2=a^2\). The \(b\) stays. So \(\dfrac{15a^4b}{5a^2}=3a^2b\).
For the second part, cube the number and multiply the index: \((3x^2)^3=3^3\times x^6=27x^6\).
Expanding brackets
Definition: expanding
To expand (or multiply out) a bracket, multiply every term inside by the term outside.
Key fact: expanding with numbers and letters
\(3(x+4)=3x+12\) and \(5(2a-3)=10a-15\). When the outside term is a letter, use the index laws: \(x(x+5)=x^2+5x\) and \(2x(3x-4)=6x^2-8x\).
Method: expand and simplify
- Multiply each term inside the first bracket by the term outside, including its sign.
- Do the same for the next bracket. A minus sign in front means multiply by a negative number.
- Collect like terms.
Worked example 3: expand and simplify
Expand and simplify \(4(2x-1)-3(x-2)\).
Expand the first bracket: \(8x-4\).
Expand the second bracket, multiplying by \(-3\): \(-3x+6\).
Combine like terms: \(8x-4-3x+6=5x+2\).
Watch out: minus signs
A very common error is writing \(-3x-6\) instead of \(-3x+6\). Multiply \(-3\) by \(-2\) and you get \(+6\).
Factorising
Definition: factorising
Factorising is the reverse of expanding. You write an expression as a product by taking out the highest common factor (HCF) of all the terms and placing the rest inside brackets.
Method: factorising
- Find the HCF of the numbers.
- Find the letters that appear in every term, using the smallest power of each.
- Write the HCF outside the bracket and divide each term by it to fill the bracket.
- Check by expanding.
Worked example 4: common number factor
Factorise \(6x+9\).
The HCF of 6 and 9 is 3. Dividing each term by 3 gives \(2x\) and \(3\). So \(6x+9=3(2x+3)\).
Worked example 5: letters and numbers
Factorise \(12x^2-8x\).
The HCF of 12 and 8 is 4. Both terms contain \(x\), and the smallest power is \(x^1\), so the HCF is \(4x\). Then \(12x^2\div4x=3x\) and \(8x\div4x=2\). So \(12x^2-8x=4x(3x-2)\).
Check by expanding: \(4x\times3x=12x^2\) and \(4x\times(-2)=-8x\). It matches.
Worked example 6: two letters
Factorise \(6ab+9a^2\).
The HCF is \(3a\). \(6ab\div3a=2b\) and \(9a^2\div3a=3a\). Answer: \(3a(2b+3a)\).
Substitution, checking and identities
Key fact: checking by substitution
You can check any simplification by substituting a number into both versions. If the values agree, the working is probably correct. Choose a number that is not 0 or 1 to avoid accidental matches.
Worked example 7: checking
Show that \(3(x+2)+2(x-1)\) simplifies to \(5x+4\) by trying \(x=3\).
The original gives \(3\times5+2\times2=15+4=19\). The answer gives \(5\times3+4=19\). The values agree, so the simplification is correct.
Definition: identity
An identity is true for every value of the letter, unlike an equation, which is true only for some values. To show an identity, expand one side and compare. For example \(3(2x+1)+3=6x+6=6(x+1)\). Some books write \(\equiv\) for an identity.
Worked example 8: algebra in context
A rectangle has sides \((x+3)\) cm and \((2x-1)\) cm. Find its perimeter in its simplest form.
Perimeter \(=2\times[(x+3)+(2x-1)]=2(3x+2)=6x+4\) cm. Algebra describes the perimeter for every value of \(x\) in one line.
Watch out: common mistakes
- Multiplying only the first term in a bracket: \(3(x+4)=3x+4\) is wrong.
- Collecting unlike terms, such as \(3x+2y=5xy\).
- Sign errors when a minus sign is in front of a bracket.
- Not taking out the highest common factor: \(12x^2-8x=2(6x^2-4x)\) is not fully factorised.
- Writing \(x^2\times x^3=x^6\); add the indices, do not multiply.
Worked example 9: algebra in a shop
A notebook costs £\(n\) and a pen costs £\(p\). Emma buys 3 notebooks and 4 pens and pays with a £20 note. Write an expression for her change and find it when \(n=2.50\) and \(p=1.20\).
Total cost \(=3n+4p\). Change \(=20-(3n+4p)=20-3n-4p\). With the numbers: \(3\times2.50+4\times1.20=7.50+4.80=12.30\), so her change is \(20-12.30=£7.70\).
Worked example 10: factorise fully
Factorise fully \(18a^2b-27ab^2\).
Find the HCF of the numbers: 9 (the largest number that divides both 18 and 27). Find the HCF of the letters: \(a\) appears as \(a^2\) and \(a\), so take \(a\); \(b\) appears as \(b\) and \(b^2\), so take \(b\). The HCF is \(9ab\). Divide each term: \(18a^2b\div9ab=2a\) and \(27ab^2\div9ab=3b\). So \(18a^2b-27ab^2=9ab(2a-3b)\). Expand to check: \(9ab\times2a=18a^2b\) and \(9ab\times3b=27ab^2\).
Worked example 11: expanding with a negative factor
Expand and simplify \(7-3(x-2)\).
Be careful: the minus sign belongs to the 3. Expand \(-3(x-2)=-3x+6\). Then \(7-3x+6=13-3x\). A common error is to write \(7-3x-6\), which gives the wrong answer \(1-3x\).
Worked example 12: substituting into a formula
The cost in pounds of hiring a bike is \(C=8+3h\), where \(h\) is the number of hours. Find the cost for 5 hours, and the number of hours that costs £26.
For 5 hours: \(C=8+3\times5=8+15=£23\). For £26: \(8+3h=26\), so \(3h=18\) and \(h=6\) hours. Substituting finds an output; solving an equation finds an input.
Note: how to choose the right tool
Look at the question word. If it says simplify, collect like terms. If it says expand, multiply out the brackets. If it says factorise, do the reverse and put the brackets back. If it says work out the value or substitute, replace each letter by its number. Writing the instruction word at the top of your working helps you to choose the correct method and avoid losing easy marks.
Watch out: signs and powers
- \(3x+4x=7x\), but \(3x\times4x=12x^2\). Adding changes the coefficient only; multiplying also changes the power.
- \((2x)^2=4x^2\), not \(2x^2\): the bracket squares the 2 as well.
- When you factorise, always check by expanding. If you do not recover the original expression, something has gone wrong.
Method: proving a number trick
- Call the starting number \(n\).
- Write an expression after every step of the trick.
- Simplify the final expression.
- If the letter \(n\) disappears, the answer is the same for every starting number.
Did you know?
The word algebra comes from the Arabic al-jabr, from a book written around the year 820 by the mathematician al-Khwarizmi in Baghdad. Today algebra is used by programmers, engineers and scientists. A spreadsheet in a shop uses algebra every time it applies a formula such as \(B2\times C2\) to find a total price.
Putting the skills together
Expanding and factorising are two directions of the same road: expanding removes brackets, factorising puts them back. Being fluent in both lets you rewrite an expression in whichever form is most useful. A factorised expression shows its factors clearly, which helps when you divide or solve equations, while an expanded expression is easier to add to other expressions or to substitute into.
Example: choosing the useful form
Show that \(4n+6\) is always even for any whole number n.
Factorising gives \(4n+6=2(2n+3)\). Because the expression is 2 multiplied by a whole number, it is a multiple of 2, so it is always even. In the expanded form this is hard to see; in the factorised form the proof takes one line. This kind of short algebraic argument is called a proof, and it is much stronger than testing a few numbers, because it works for every value of n at once.
Did you know?
The word “algebra” comes from the title of a book written in Baghdad around the year 820 by Muhammad ibn Musa al-Khwarizmi. His name also gave us the word “algorithm”, which today describes the step-by-step instructions that computers follow.
Otto's tip
Expand to check a factorised answer. If expanding your brackets brings you back to the start, you are right. Factorising and expanding are a pair of opposite moves, like going up and down a staircase.
How to use this Year 8 algebra lesson
First, read the key points so that you know where the Year 8 algebra lesson is going. Then work through each definition and worked example, and cover the solution before you read it. After that, practise with the algebra: expressions 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 algebra: expressions 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 8 algebra lesson 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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