Algebra | ICSE Class 8 Maths Notes

This note covers algebraic expressions, operations on expressions, identities, simple factorisation, division, linear equations in one variable, contextual problems, linear relationships, pairs of linear equations and inequalities in one variable.

What are algebraic expressions, terms and coefficients?

Algebra uses letters to represent numbers and express relationships. A variable is a symbol whose value can change. Here, x and y are variables representing numbers. A constant has a fixed value. Combining variables and constants through operations produces an algebraic expression, such as x + 3 or 2y − 5.

The signs + and − indicate addition and subtraction; × indicates multiplication, and ÷ or a fraction bar indicates division. The sign = means “is equal to”. Letters written together indicate multiplication: xy = x × y. Brackets keep an expression together as a unit.

A term is a part added to form an expression; a subtraction can be read as addition of a negative term. A factor is a number or expression multiplied to form a product. In 5xy + 3x, the terms are 5xy and 3x; factors of 5xy include 5, x and y.

The numerical coefficient is the number multiplying the variable part of a term. The coefficients of 5xy and 3x are 5 and 3. A power records repeated multiplication: x² means x × x, and x³ means x × x × x. The raised number is the exponent.

How are expressions classified?

A monomial has one term, a binomial has two terms, and a trinomial has three terms. A polynomial contains terms with non-zero coefficients and non-negative integer exponents of variables. Integers are whole numbers and their negatives, including zero; non-negative means zero or positive.

Count terms after recognising which operations join them. Multiplication inside a term does not create another term. Thus 3x² is a monomial, while 4xy + 7 is a binomial. The letter part and its powers matter when deciding which terms can be combined.

How do like terms help with addition and subtraction?

Like terms contain the same variables raised to the same powers. Their numerical coefficients can differ. In contrast, unlike terms have different variable parts or powers. Combining like terms changes their coefficients while retaining the common variable part.

For addition, arrange corresponding terms together. Terms without a matching like term remain in the answer. A constant can combine with another constant. Keep each sign attached to its term, especially when arranging expressions in rows.

Worked example 1. Add 7x² − 4x + 5 and 9x − 10.

Answer: 7x² − 4x + 5 + 9x − 10 = 7x² + (−4 + 9)x + (5 − 10) = 7x² + 5x − 5. The square term remains because the second expression has no matching square term.

What changes when an expression is subtracted?

The additive inverse of a term is the term with the opposite sign, so that their sum is zero. Subtracting an expression means adding the additive inverse of every term in it. A minus sign before a bracket therefore affects the whole bracket.

Worked example 2. Subtract 5x² − 4y² + 6y − 3 from 7x² − 4xy + 8y² + 5x − 3y.

Answer: Write the second expression first, then subtract the bracket containing the first. This gives 7x² − 4xy + 8y² + 5x − 3y − 5x² + 4y² − 6y + 3. Combining like terms gives 2x² − 4xy + 12y² + 5x − 9y + 3.

The wording “subtract from” fixes the order. Reversing that order changes the calculation. In this example, subtracting the negative constant produces a positive constant, while subtracting the positive y term produces another negative y term.

How do you multiply monomials and expand brackets?

To multiply monomials, multiply the numerical coefficients and then multiply the variable factors. Repeated occurrences of the same variable are collected into a power. The sign of the coefficient must be handled along with its numerical value.

Worked example 3. Multiply 5x by 4x².

Answer: 5x × 4x² = (5 × 4)(x × x²) = 20x³. The coefficient is 20, and the three factors equal to x give x³.

The same method handles negative coefficients. For instance, 5x × (−3y) = −15xy. Here the variables differ, so their product is written xy rather than a power of either one. Multiplication of monomials produces a monomial.

Property: distributive multiplication

The distributive property means multiplying each term inside a bracket by the factor outside it. An expansion writes a product involving brackets as a sum of terms. Each term must receive the outside multiplier, including terms with negative signs.

Worked example 4. Expand 3x(5y + 2).

Answer: 3x(5y + 2) = (3x × 5y) + (3x × 2) = 15xy + 6x. Both terms inside the bracket are multiplied by 3x.

With a negative outside factor, the same rule applies: (−3x)(−5y + 2) = 15xy − 6x. Do the multiplication before combining terms. Addition and multiplication have different effects: multiplying variable factors can change powers, whereas adding like terms changes coefficients.

Evaluation means finding an expression's numerical value after replacing its variables by given numbers. For x(x − 3) + 2 at x = 1, expansion gives x² − 3x + 2. Substitution then gives 1 − 3 + 2 = 0.

How do you multiply two binomials?

Use the distributive property twice when multiplying two binomials. Each term in the first bracket multiplies each term in the second. There are four initial products, though combining like terms may reduce the number of terms in the final expression.

Let a and b represent numbers. In (3a + 4b)(2a + 3b), first multiply the whole second bracket by 3a, then by 4b. This gives 6a² + 9ab + 8ba + 12b². Since ba = ab, the result is 6a² + 17ab + 12b².

Worked example 5. Multiply (x − 4) and (2x + 3).

Answer: (x − 4)(2x + 3) = x(2x + 3) − 4(2x + 3) = 2x² + 3x − 8x − 12 = 2x² − 5x − 12.

How can you check that every product is included?

1. Take the first term of the first bracket and multiply it by each term of the second bracket.
2. Repeat with the second term of the first bracket, retaining its sign.
3. Write all the products before trying to combine any like terms.
4. Collect like terms and check that no original term has been skipped.

The commutative property of multiplication allows factors to change order without changing their product. It explains why ab and ba combine. It does not make a² and ab like terms: those expressions have different variable factors.

This method extends to larger polynomials. Each term still multiplies every term in the other expression. Simplification comes after this distribution, so it should not hide an omitted product or an incorrect sign.

What identities follow from multiplying binomials?

An identity is an equality true for all permitted values of its variables. An equation to be solved instead asks which values make a particular equality true. Identities provide general patterns for expansion and for reversing an expansion through factorisation.

Identity: square of a sum

(a + b)² = a² + 2ab + b². Here a and b represent numbers, and the square applies to the whole bracket. To prove the identity, write the square as the product of two identical brackets.

1. Write (a + b)² = (a + b)(a + b).
2. Distribute the first bracket to obtain a(a + b) + b(a + b).
3. Expand both products to obtain a² + ab + ba + b².
4. Combine ab and ba to obtain a² + 2ab + b².

Identity: square of a difference

(a − b)² = a² − 2ab + b². Multiplying (a − b)(a − b) gives a² − ab − ba + b². Combining the two negative middle terms proves the identity. The last term is positive because it comes from multiplying −b by −b.

A perfect-square expression is one that can be written as the square of an expression. To recognise either square identity, check the first term, last term and middle term. Matching the two square terms alone does not establish the pattern.

Identity: difference of squares

a² − b² = (a − b)(a + b). Expanding the right side gives a² + ab − ba − b². The middle terms cancel because ab = ba. The result contains the difference of two squares.

The brackets in this identity have the same two parts but opposite signs between them. This is different from squaring a difference, which retains a middle term. Read the original operation carefully before selecting an identity.

Note: A numerical check can help detect a mistake, but the distributive proofs explain why these identities hold generally. For factorisation, read the same identities from the expanded expression towards the product of brackets.

How does taking a common factor reverse expansion?

Factorisation means writing an expression as a product of factors. A common factor occurs in each of the terms being considered. Removing it from the terms and placing it outside a bracket reverses distributive multiplication.

For example, 2x + 4 = 2(x + 2). Both original terms contain a factor 2. Similarly, 5xy + 10x = 5x(y + 2), because 5x is common to both terms. Multiplying the brackets out checks the factorisation.

Worked example 6. Factorise 12a²b + 15ab².

Answer: The common factors are 3, a and b. Write 12a²b = 3ab × 4a and 15ab² = 3ab × 5b. Therefore, 12a²b + 15ab² = 3ab(4a + 5b).

How can regrouping reveal a common factor?

Regrouping means rearranging terms into groups that can be factorised. Sometimes a useful factor is shared within separate groups even though no non-trivial factor is common to every original term. A non-trivial factor here means a factor other than 1.

Worked example 7. Factorise 2xy + 2y + 3x + 3.

Answer: Group the expression as (2xy + 2y) + (3x + 3). Taking common factors gives 2y(x + 1) + 3(x + 1). The bracket x + 1 is now common, so the answer is (x + 1)(2y + 3).

Not every rearrangement leads to the desired common bracket. Choose groups by inspecting their factors. When taking out a factor leaves no variable or numerical factor other than 1, retain that 1 inside the bracket, as in 2y(x + 1).

How do identities help you factorise simple expressions?

Compare the entire expression with an identity before writing its factors. Identify the quantities being squared, then check the middle term if one is present. The quantities corresponding to a and b may themselves be terms such as 2y or 7p, where p is another variable.

Worked example 8. Factorise x² + 8x + 16.

Answer: Write x² + 8x + 16 = x² + 2 × x × 4 + 4². This is the square-of-a-sum pattern, so the factorised expression is (x + 4)².

For a negative middle term, 4y² − 12y + 9 = (2y)² − 2 × (2y) × 3 + 3² = (2y − 3)². For a difference of squares, 49p² − 36 = (7p)² − 6² = (7p − 6)(7p + 6).

Identity: factors with a shared variable

(x + a)(x + b) = x² + (a + b)x + ab. Here x is the variable and a and b are fixed numbers. Expanding the brackets gives x² + ax + bx + ab, and combining the middle terms proves the identity.

To reverse this pattern, find two numbers whose sum is the coefficient of x and whose product is the constant term. Both conditions matter. A pair that gives the correct product but the wrong sum does not give the required factors.

Worked example 9. Factorise x² + 5x + 6.

Answer: The numbers 2 and 3 have product 6 and sum 5. Split the middle term: x² + 2x + 3x + 6 = x(x + 2) + 3(x + 2) = (x + 2)(x + 3).

Signs are part of the search. In y² − 7y + 12, the suitable numbers are −3 and −4. Hence y² − 7y + 12 = (y − 3)(y − 4). Their product is positive while their sum is negative.

How do you divide algebraic expressions?

Division reverses multiplication. The expression being divided is the dividend, the expression dividing it is the divisor, and the result is the quotient. In an exact division, dividend = divisor × quotient, with no remainder left over.

A divisor must be non-zero. The sign ≠ means “is not equal to”. When division involves variables, retain any restriction needed to prevent the original divisor from becoming zero. Simplifying an expression does not remove that restriction.

Worked example 10. Divide 6x³ by 2x, where x ≠ 0.

Answer: Write 6x³ = (2x)(3x²). Cancelling the common non-zero factor 2x gives 6x³ ÷ 2x = 3x². Multiplication checks the result: 2x × 3x² = 6x³.

When should you divide term by term?

When a polynomial is divided by a monomial, divide each term by that monomial and combine the results. Alternatively, factor out the divisor from the entire dividend. For instance, (5x² + 20x) ÷ 5x = x + 4 for x ≠ 0.

For division by a polynomial, factorise and look for matching factors. In a fraction, the numerator is the expression above the fraction bar, and the denominator is below it. Cancel common factors of the whole numerator and denominator, rather than individual terms of a sum.

Worked example 11. Divide 7x² + 14x by x + 2, where x ≠ −2.

Answer: First factorise 7x² + 14x = 7x(x + 2). Then 7x(x + 2) ÷ (x + 2) = 7x. The restriction ensures that the cancelled bracket is non-zero.

The remainder is what is left after division. These examples have remainder zero. Their checking rule is multiplication: multiply the quotient by the divisor and recover the original dividend.

How are linear equations in one variable solved?

An equation states that two expressions are equal. Its left-hand side, abbreviated LHS, is before the equals sign; its right-hand side, abbreviated RHS, is after it. A solution is a value that makes the equality true.

A linear equation in one variable contains one variable with highest power 1 in its linear expressions. It may have variable terms on both sides. Solving it means using operations that preserve equality until the value of the variable is obtained.

Property: preserving equality

Add or subtract the same quantity on both sides. Multiply or divide both sides by the same non-zero number. These operations keep the equation balanced. Transposition is a shortened way of recording the corresponding operation on both sides.

Worked example 12. Solve 2x − 3 = x + 2.

Answer: Add 3 to both sides to obtain 2x = x + 5. Subtract x from both sides to obtain x = 5. Checking in the original equation gives 2 × 5 − 3 = 7 and 5 + 2 = 7, so both sides agree.

A rational number can be written as a fraction of integers with a non-zero denominator. Simple rational coefficients can be handled by multiplying the whole equation by a suitable number to remove fractions. Every term on both sides must be included.

Worked example 13. Solve 5x + 7/2 = 3x/2 − 14.

Answer: Multiply both sides by 2: 10x + 7 = 3x − 28. Subtract 3x and then 7 from both sides to get 7x = −35. Divide by 7 to obtain x = −5.

Substitution replaces a variable by a value or an equal expression. Check solutions by substitution in the original equation. This catches errors that may be hidden if the check uses an already incorrect intermediate equation.

How do you turn a word problem into an equation?

A contextual problem gives relationships between known and unknown quantities. Choose a variable for an unknown quantity and state what it represents, including its unit when relevant. Then express the other quantities using the same variable wherever possible.

Translate the relationship into an equation before solving. Words indicating multiplication, division, addition or subtraction determine the operations. After solving, return to the original setting and state the requested quantities, rather than leaving an unexplained value of a letter.

Worked example 14. Half the perimeter of a rectangular garden is 36 m. Its length is 4 m more than its width. Find its dimensions. Here m denotes metres, a unit of length, and perimeter means the total length around the boundary.

Answer: Let x metres be the width; the length is (x + 4) metres. Half a rectangle's perimeter is length plus width, so x + (x + 4) = 36. Thus 2x + 4 = 36, 2x = 32 and x = 16. The width is 16 m and the length is 20 m.

How do you verify the interpretation?

The dimensions differ by 4 m, as required, and their sum is 36 m, the stated half-perimeter. Both conditions are needed. Solving an equation correctly is insufficient if the equation did not express the original information correctly.

Keep a quantity and its unit distinct from a variable's algebraic role. In this example, x records the numerical width in metres. The final answer names width and length separately, so the reader can tell which result answers each part of the question.

How can a linear equation represent two related quantities?

A linear relationship between two variables can be written y = mx + b. Here x and y represent the related quantities, m is a fixed multiplier and b is a fixed added amount. In this formula, m is a number, rather than the abbreviation for metres.

When b = 0, the relationship becomes y = mx. For x ≠ 0, this can also be written y/x = m. This is a proportional relationship: the ratio of y to x stays constant. The general linear relationship allows a fixed added amount as well.

Bivariate data consist of paired measurements of two quantities, such as a student's arm span and height. Arm span is the distance between the fingertips with arms stretched out. A model is a mathematical relationship used to describe the association between the quantities.

How should the relationship be interpreted?

Identify what each variable measures before interpreting the equation. The multiplier and the fixed added amount need meanings in that setting. Fitting a model and assessing its fit are done informally, by considering how well the relationship describes the data.

The fit of a model describes how closely it agrees with the observations. A relationship proposed to describe measurements should be interpreted in their context. The arm-span and height example concerns paired measurements; it does not supply a particular numerical equation.

Expressions, equations and models perform different tasks. An expression represents a quantity; an equation states a relationship; a model uses such a relationship to describe a situation. Recognising the role of each helps when moving from words or measurements to algebra.

How are pairs of linear equations solved and interpreted?

A pair of linear equations in two variables gives two conditions involving the same two unknowns. A common solution must satisfy both equations. In the substitution method, express one unknown in terms of the other and substitute into the second equation.

Worked example 15. Akhila plays Hoopla, a ring-throwing game, half as many times as she rides the Giant Wheel. Each ride costs ₹3 and each game costs ₹4. She spends ₹20 altogether. Find both counts. The symbol ₹ denotes rupees.

Answer: Let x be the number of rides and y the number of games. Then y = x/2 and 3x + 4y = 20. Substituting gives 3x + 4(x/2) = 20, so 5x = 20 and x = 4. Hence y = 2: four rides and two games.

The elimination method removes one unknown by adding or subtracting suitable multiples of the equations. After finding the remaining unknown, substitute back to obtain the other. In either method, check the resulting pair in both original equations.

What do the corresponding lines show?

A graph represents a relationship by points. The coordinate axes are the horizontal and vertical reference lines. An ordered pair (x, y) records the horizontal coordinate first and the vertical coordinate second. A linear equation in two variables is represented by a straight line.

The horizontal and vertical axes are labelled X and Y. Points A(0, 2) and B(6, 0) lie on one line; P(0, −4) and Q(3, −2) lie on the other. A, B, P and Q name the plotted points. The two lines meet at B(6, 0).

See Fig. 3.1 in your NCERT textbook

The equations for these lines are x + 3y = 6 and 2x − 3y = 12. Their common solution is x = 6, y = 0. A unique solution means exactly one common solution, rather than a separate answer chosen independently for each equation.

How do you solve inequalities in one variable?

An inequality compares quantities using < for less than, > for greater than, ≤ for less than or equal to, or ≥ for greater than or equal to. A strict inequality uses < or >, so equality is excluded.

A solution makes the comparison true. The solution set is the collection of all permitted solutions. State which numbers are allowed: natural numbers are the positive counting numbers, integers include zero and negatives, and real numbers are the numbers represented on the number line.

Property: operations on inequalities

Add or subtract the same number on both sides without changing the comparison sign. Multiplying or dividing both sides by the same positive number also preserves it. Multiplying or dividing by a negative number reverses the sign.

Worked example 16. Solve 4x + 3 < 6x + 7 for real x.

Answer: Subtract 6x and then 3 from both sides to get −2x < 4. Dividing by −2 reverses the sign, giving x > −2. Every real number greater than −2 is a solution; −2 itself is excluded.

The permitted number system affects how an answer is stated. For 30x < 200, division by 30 gives x < 20/3. If x is a natural number, the solutions are 1, 2, 3, 4, 5 and 6. Integer solutions also include zero and all negative integers.

How does a number line display the answer?

A number line places numbers in order on a straight line. An open circle at a boundary means the boundary is excluded. For 7x + 3 < 5x + 9, subtraction gives 2x < 6, so x < 3.

The number line has labelled integer marks from −4 to 6. An open circle is drawn at 3, and a blue line with an arrow extends to the left, showing the numbers less than 3.

See Fig. 5.1 in your NCERT textbook

Do not replace a strict inequality by one that includes equality. The boundary is not part of this solution. Likewise, finding one number that works does not describe the whole solution set when the question asks for all solutions.

Glossary

• Variable — A symbol representing a number whose value can change in an expression or relationship.
• Constant — A number with a fixed value in the expression being considered.
• Term — A part added to form an expression, with its own sign retained.
• Coefficient — The numerical factor multiplying the variable part of an algebraic term.
• Exponent — The raised number indicating the power to which a variable or number is taken.
• Like terms — Terms containing the same variables raised to the same respective powers.
• Polynomial — An expression formed from terms with non-zero coefficients and non-negative integer exponents of variables.
• Identity — An equality that holds for all permitted values of its variables.
• Factorisation — Writing an algebraic expression as a product of numerical or algebraic factors.
• Quotient — The result obtained when one number or expression is divided by another.
• Linear equation — An equation formed from linear expressions, with variables occurring to the first power.
• Substitution — Replacing a variable by a given value or by an equal expression.
• Elimination — Removing an unknown by adding or subtracting suitable multiples of two equations.
• Inequality — A comparison expressing that quantities are less than, greater than, or possibly equal to one another.
• Solution set — The collection of all permitted values that make an equation or inequality true.

Common errors and misconceptions

• Misconception: 2 + x can be simplified to 2x. Correct: The first expression adds 2 to x; the second multiplies x by 2. These are different operations and do not give a general identity.
• Misconception: 7x + y can be written as 7xy. Correct: Unlike terms cannot be combined that way. The expression 7xy is a product, while 7x + y is a sum.
• Misconception: Subtracting a bracket changes only its first sign. Correct: Subtract each term, or add the additive inverse of the whole expression, before collecting like terms.
• Misconception: Squaring a sum requires only squaring its two terms. Correct: Include the middle term 2ab in (a + b)², because both cross-products occur during multiplication.
• Misconception: Matching the constant product is sufficient when choosing two factors. Correct: The chosen numbers must also have the required sum, including the correct signs.
• Misconception: Terms of a sum can be cancelled separately against a divisor. Correct: Factorise first and cancel common non-zero factors of the entire numerator and denominator.
• Misconception: Dividing an inequality by a negative number leaves its sign unchanged. Correct: Reverse the comparison sign, and preserve whether the boundary is included or excluded.

Exam-style questions with model answers

Q1. In 5xy + 3x, where x and y are variables, identify the terms and their numerical coefficients. [2 marks]

1. The terms are 5xy and 3x, since these two parts are added to form the expression.
2. The numerical coefficient of 5xy is 5, and the numerical coefficient of 3x is 3.

Q2. Expand and simplify (x − 4)(2x + 3), where x is a variable. Explain how the like terms combine. [3 marks]

1. Distribute the first bracket across the second: (x − 4)(2x + 3) = x(2x + 3) − 4(2x + 3). This ensures that both terms in each bracket are included.
2. Multiply term by term to obtain 2x² + 3x − 8x − 12. The negative multiplier gives negative products in the second pair.
3. Combine the like terms 3x and −8x to give −5x. The simplified product is 2x² − 5x − 12.

Q3. Prove the identity (a − b)² = a² − 2ab + b² by multiplication, where a and b represent numbers. [4 marks]

1. A square means multiplication by the same expression, so write (a − b)² = (a − b)(a − b).
2. Apply the distributive property to obtain a(a − b) − b(a − b), keeping the negative sign with the second multiplier.
3. Expand both products to get a² − ab − ba + b². The last term is positive because two negative factors are multiplied.
4. Since ba = ab, the middle terms combine to −2ab. Hence the expression equals a² − 2ab + b², proving the identity.

Q4. Factorise x² + 5x + 6, where x is a variable, and check the factors by multiplication. [4 marks]

1. Use the pattern in which the two constant parts of the factors have product 6 and sum 5.
2. The numbers 2 and 3 satisfy both conditions. Split the middle term to write x² + 2x + 3x + 6.
3. Group terms to obtain x(x + 2) + 3(x + 2), then take out the common bracket to get (x + 2)(x + 3).
4. Multiplying back gives x² + 3x + 2x + 6 = x² + 5x + 6, so the factorisation is checked.

Q5. A rectangular garden has half-perimeter 36 m, and its length is 4 m greater than its width. Here m means metres. Using half-perimeter = length + width, form a linear equation, find both dimensions and check them. [5 marks]

1. Let x be the numerical width in metres. The length is then x + 4 metres because it exceeds the width by 4 metres.
2. The stated half-perimeter equals length plus width. Therefore, the equation representing the information is x + (x + 4) = 36.
3. Combine like terms to obtain 2x + 4 = 36. Subtract 4 from both sides, giving 2x = 32.
4. Divide both sides by 2 to obtain x = 16. Thus the width is 16 m and the length is 16 + 4 = 20 m.
5. Check both conditions: 20 − 16 = 4, and 20 + 16 = 36. The dimensions have the required difference and half-perimeter.

Q6. Akhila plays Hoopla, a ring-throwing game, half as many times as she rides the Giant Wheel. A ride costs ₹3 and a game costs ₹4; ₹ denotes rupees. Her total spending is ₹20. Form two equations, solve them and check both counts. [5 marks]

1. Let x represent the number of rides and y the number of games. The count relationship is y = x/2 because the games are half as numerous as the rides.
2. The rides cost 3x rupees and the games cost 4y rupees. The total spending condition therefore gives the equation 3x + 4y = 20.
3. Substitute y = x/2 into the spending equation: 3x + 4(x/2) = 20. Simplifying gives 3x + 2x = 20, or 5x = 20.
4. Divide by 5 to get x = 4, and substitute into y = x/2 to get y = 2. She takes four rides and plays twice.
5. The game count 2 is half the ride count 4. The spending is 3 × 4 + 4 × 2 = 20 rupees, so both conditions hold.

Q7. Solve 4x + 3 < 6x + 7 for real x, meaning x may be any number on the number line. Explain the change of sign and whether the boundary is included. [3 marks]

1. Subtract 6x from both sides, then subtract 3. This gives −2x < 4 without changing the comparison sign.
2. Divide both sides by −2. Division by a negative number reverses the comparison, so the solution is x > −2.
3. The solution includes every real number greater than −2. It excludes −2 itself because the inequality is strict and does not allow equality.

Q8. Divide 7x² + 14x by x + 2, where x is a variable and x ≠ −2. Show the factorisation, cancellation and multiplication check, and explain the restriction. [4 marks]

1. The common factor of 7x² and 14x is 7x. Hence the dividend can be written as 7x(x + 2).
2. The division becomes 7x(x + 2) ÷ (x + 2). Cancel the whole common bracket to obtain the quotient 7x.
3. The restriction x ≠ −2 ensures that x + 2 is non-zero, so the original division and the cancellation are permitted.
4. Multiplying the quotient by the divisor gives 7x(x + 2) = 7x² + 14x. This recovers the dividend and verifies the answer.

Key takeaways

• Identify terms, coefficients and variable powers before choosing an operation; factors are multiplied, while terms are added to form expressions.
• Add and subtract like terms by combining their coefficients, and retain the signs of terms throughout the calculation.
• Distribute every term across the other bracket when multiplying polynomials, then combine any like terms produced.
• Square identities include a middle term; the difference-of-squares identity gives two brackets whose signs differ.
• Factorisation reverses expansion through common factors, suitable regrouping or recognition of a matching algebraic identity.
• In division, cancel common non-zero factors after factorisation and retain restrictions imposed by the original divisor.
• Form equations from the stated relationships, perform equal operations on both sides and check solutions in the original conditions.
• Pairs of equations require common solutions; inequalities require attention to the permitted numbers, boundary and direction of comparison.

Test yourself

What makes two algebraic terms like terms?

They contain the same variables raised to the same respective powers; their numerical coefficients can differ.

For variables x and y, why does 3x(5y + 2) contain two products?

The outside factor 3x multiplies each term inside the bracket, giving 15xy and 6x, which are then added.

For numbers a and b, what middle term occurs in (a − b)²?

The middle term is −2ab, obtained by combining the two negative products −ab and −ba.

What two conditions must the constant parts of the factors of x² + 5x + 6 satisfy?

Their sum must be 5 and their product must be 6. The numbers 2 and 3 meet both conditions.

Why is x ≠ −2 required when dividing 7x² + 14x by x + 2?

At x = −2, the original divisor x + 2 equals zero, so the division is not defined.

What does a point common to two straight-line graphs represent?

Its coordinates give values of the two variables that satisfy both equations, so it represents a common solution.

What happens when both sides of an inequality are divided by a negative number?

The comparison sign reverses. Greater than becomes less than, and greater than or equal to becomes less than or equal to.

If x is a natural number, which values solve 30x < 200?

Dividing by 30 gives x < 20/3. The permitted natural-number solutions are 1, 2, 3, 4, 5 and 6.