Expansions | ICSE Class 9 Maths Notes

This note covers expansion of algebraic expressions, squares of sums and differences, products with a common term, squares of three-term expressions, cubes of sums and differences, geometrical interpretations, numerical calculations and checks on signs and coefficients.

What does it mean to expand an algebraic expression?

How should the notation be read?

An algebraic expression combines numbers and letters through mathematical operations. A variable is a letter representing a number. Throughout the identities below, a, b, c, x and y represent real numbers, meaning numbers on the number line, unless a geometrical interpretation requires positive lengths.

Writing ab means a multiplied by b; the multiplication sign × has the same meaning. The superscripts in a² and a³ indicate powers: a² means a × a, while a³ means a × a × a. Brackets group an expression to be treated together.

A term is a part separated from other parts by addition or subtraction, with its sign retained. A binomial has two unlike terms; a trinomial has three. A coefficient is the numerical multiplier of a variable part.

Like terms have identical variable parts, including their powers. Combining them means adding their coefficients. A factor is an expression being multiplied by another expression. A product is the result of multiplication. These distinctions help explain what changes during expansion.

Definition: Expansion rewrites a product or a power of a bracketed expression as a sum of terms, combining like terms where possible. The value of the expression is preserved.

Why does an identity work for different values?

An equation states that two expressions are equal, using the sign =. An algebraic identity is an equation true for all values of its variables. An equation need not be an identity: x² − 1 = 24 holds for x = 5 or x = −5.

The symbol − means subtraction or a negative sign, according to its position. The distributive property states that a(b + c) = ab + ac. It supplies the reason for multiplying each term in one bracket by each term in another bracket.

A calculation with selected values can check an identity for those values. A derivation using distribution establishes the result generally. When applying an identity, first identify the complete expressions replacing its letters, then carry out every indicated product and power.

How is the square of a sum expanded?

Identity: square of a binomial sum

(a + b)² = a² + 2ab + b². The square of a sum contains the square of each term and twice their product. The term 2ab is called a cross term because it combines the two different parts of the binomial.

To derive the identity, write (a + b)² as (a + b)(a + b). Distribution gives a² + ab + ba + b². Since ab and ba are equal, their sum is 2ab. This accounts for the middle term rather than merely asking you to remember it.

A square of side a + b is divided into a yellow square labelled a², a purple square labelled b² and two green rectangles labelled ab. Its sides are split into lengths a and b.

See Fig. 4.2 in your NCERT textbook

In the diagram, area measures the surface enclosed by a shape. A square's area is its side multiplied by itself. A rectangle's area is the product of its adjacent side lengths. Adding the four component areas explains the square identity for positive lengths.

How are whole terms substituted?

Worked example 1. Expand (5x + 2y)², where x and y are real-number variables.

Answer: Take a = 5x and b = 2y. Then (5x + 2y)² = (5x)² + 2(5x)(2y) + (2y)² = 25x² + 20xy + 4y².

The coefficient and variable are both squared in each square term. In the middle term, multiply 2, 5 and 2 to obtain 20, and multiply x by y to obtain xy.

The same formula applies when a or b is negative. For a = −2 and b = −3, the bracket equals −5, whose square is 25. The expanded form gives 4 + 12 + 9 = 25, agreeing with the direct calculation.

This numerical check illustrates the identity; distribution explains its general validity. The geometric picture uses lengths, so negative numbers should be handled through the algebraic reasoning rather than interpreted as negative side lengths.

How is the square of a difference expanded?

Identity: square of a binomial difference

(a − b)² = a² − 2ab + b². Replace b by −b in the sum identity to obtain this expression. The product 2a(−b) becomes −2ab, while (−b)² becomes b² because multiplying two negative factors gives a positive product.

Direct multiplication gives the same result: (a − b)(a − b) = a² − ab − ba + b². Combining the two middle terms gives −2ab. The final square term has a plus sign even though the original binomial contains subtraction.

The outer square has side a. A smaller upper-left square is labelled (a − b)². A red rectangle on the right is labelled ab, and the yellow lower-left rectangle is labelled b(a − b).

See Fig. 4.3 in your NCERT textbook

For this picture, a is greater than b and both are positive lengths. Subtracting the two rectangular areas from the outer square gives (a − b)² = a² − ab − b(a − b). Distributing the final subtraction produces a² − 2ab + b².

How does the identity simplify arithmetic?

Worked example 2. Find 29² using the square of a difference.

Answer: Write 29 = 30 − 1. Then 29² = (30 − 1)² = 30² − 2 × 30 × 1 + 1² = 900 − 60 + 1 = 841.

The number subtracted inside the bracket is squared as well as used in the middle product. Retaining that final square gives the correct total.

Choose the difference form when a nearby convenient number is larger than the given number. This choice makes the component calculations easier. It does not change the identity or permit any term to be left out.

Note: The sign before b² in the expansion is positive. The middle term is −2ab; its numerical value still depends on the signs of the values assigned to a and b.

How are products of the form (x ± a)(x ± b) expanded?

Identity: product with a common first term

The symbol ± means plus or minus. In the family (x ± a)(x ± b), each bracket may have a plus or a minus sign. Write the signs explicitly before calculating so that the product and the combined middle term are unambiguous.

(x + a)(x + b) = x² + (a + b)x + ab. Distribution first gives x² + bx + ax + ab. The middle terms contain the same variable x, so their coefficients combine to give a + b.

The term ab is the constant term with respect to x when a and b are fixed numbers. It is obtained by multiplying the parts independent of x. Both the sum a + b and the product ab matter; they play different roles.

How do the signs affect the answer?

Worked example 3. Expand (x + 3)(x + 4).

Answer: The four products give x² + 4x + 3x + 12. Combining like terms gives x² + 7x + 12. The coefficient of x is 3 + 4, and the constant term is 3 × 4.

Worked example 4. Expand (x − 2)(x − 3).

Answer: Distribute to obtain x² − 3x − 2x + 6. Therefore, (x − 2)(x − 3) = x² − 5x + 6. The two negative constant parts multiply to give positive 6.

A useful special case is (x + a)(x − a) = x² − a². The middle products cancel because they are equal in size and opposite in sign. This is the difference-of-squares identity, used here to expand a product.

Compare the two brackets before choosing a formula. A repeated bracket gives a square, while equal first terms with opposite second terms give a difference of squares. Different second terms require their signed sum and their signed product.

How is the square of three added terms expanded?

Identity: square of a trinomial sum

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca. The expansion contains three individual squares and three doubled pairwise products. Here pairwise means taking two different terms together, once for each possible pair.

One way to derive it is to group b + c as a single expression. Apply the sum-square identity to a + (b + c), giving a² + 2a(b + c) + (b + c)². Now expand the product and the remaining square.

This gives a² + 2ab + 2ac + b² + 2bc + c². Rearranging the terms gives the displayed identity, since ac and ca are the same product. Grouping changes the organisation of the calculation without changing its value.

A square is divided into nine coloured regions. Its top and left edges are each split into lengths labelled a, b and c. Matching divisions form three squares and six rectangles.

See Fig. 4.4 in your NCERT textbook

The three square regions have areas a², b² and c². The rectangles occur in equal-area pairs: two have area ab, two have area bc and two have area ca. Adding these areas gives the complete expression for the outer square.

How can three parts simplify a numerical square?

Worked example 5. Calculate 119² by writing 119 as 100 + 10 + 9.

Answer: 119² = 100² + 10² + 9² + 2(100)(10) + 2(100)(9) + 2(10)(9).

Thus 119² = 10000 + 100 + 81 + 2000 + 1800 + 180 = 14161. All three individual squares and all three doubled products contribute to the answer.

For a systematic check, list the three squares first, then the products of the first and second, second and third, and third and first terms. This order makes a missing pair easier to spot without relying on the appearance of the final answer.

How do minus signs change a three-term square?

How are signed terms handled?

A signed term is a term considered together with its positive or negative sign. For a square such as (a − b + c)², apply the three-term identity to a, −b and c. Each sign belongs to the complete term being substituted.

Squaring the individual signed terms gives a², b² and c². The doubled products are 2a(−b), 2(−b)c and 2ca. Consequently, the two products involving −b are negative in the written expansion, while the product involving a and c is positive.

In the last row, the product of the two negative terms is positive. This explains the plus sign before 2bc. It is safer to multiply the signed terms than to assume that every cross term becomes negative when subtraction appears.

How is a mixed-sign expression expanded?

Worked example 6. Expand (3x − 2y + 4z)², where z is another real-number variable.

Answer: The individual squares are (3x)² = 9x², (−2y)² = 4y² and (4z)² = 16z².

The doubled products are 2(3x)(−2y) = −12xy, 2(−2y)(4z) = −16yz and 2(4z)(3x) = 24xz.

Combining these gives (3x − 2y + 4z)² = 9x² + 4y² + 16z² − 12xy − 16yz + 24xz.

Terms such as xy, yz and xz are different variable parts, so they are not like terms and cannot be combined into one term. Retain them separately. Check each product against the two original terms that generated it.

The plus signs before the individual squares come from squaring signed quantities. They do not remove the need to examine the cross terms. Keeping those two stages separate makes both the coefficients and the signs easier to verify.

How is the cube of a sum expanded?

Identity: cube of a binomial sum

(a + b)³ = a³ + 3a²b + 3ab² + b³. The cube of an expression means the product of three equal factors, each equal to that expression. Unlike a square, this expansion contains cubes and two kinds of mixed product.

Begin with (a + b)³ = (a + b)(a² + 2ab + b²). Multiply the three terms in the second bracket first by a and then by b. The resulting terms explain both coefficients of 3.

1. Multiplication by a gives a³ + 2a²b + ab².
2. Multiplication by b gives a²b + 2ab² + b³.
3. Combine 2a²b with a²b to obtain 3a²b.
4. Combine ab² with 2ab² to obtain 3ab² and retain both cubes.

The figure separates a cube into two purple cubes, three yellow cuboids and three green cuboids. The grouped pieces are labelled a³, 3a²b, 3ab² and b³, with edges marked a or b.

See Fig. 4.10 in your NCERT textbook

A cuboid is a rectangular solid; its volume, the space it occupies, is the product of its three dimensions. Three cuboids have dimensions a, a and b, and three have dimensions a, b and b. Together with the cubes, they account for the identity.

How are larger terms substituted into the cube identity?

Worked example 7. Expand (p + 2q)³, where p and q are real-number variables.

Answer: Substitute a = p and b = 2q. Then (p + 2q)³ = p³ + 3p²(2q) + 3p(2q)² + (2q)³ = p³ + 6p²q + 12pq² + 8q³.

In the third term, squaring 2q gives 4q² before multiplication by 3p. In the final term, cubing 2q gives 8q³. The coefficients therefore arise from different operations.

The mixed terms p²q and pq² are unlike terms. One contains two factors of p, while the other contains two factors of q. Their coefficients cannot be added as though their variable parts were identical.

How is the cube of a difference expanded?

Identity: cube of a binomial difference

(a − b)³ = a³ − 3a²b + 3ab² − b³. Replace b by −b in the sum-cube identity. The powers of the substituted negative term explain why the displayed signs alternate between plus and minus.

The term with one factor of −b is −3a²b. The term with two factors of −b is +3ab². The final term has three factors of −b, giving −b³. This is why the last term differs in sign from the final term of a difference square.

The symbols show the algebraic sign pattern. If a or b is itself assigned a negative value, evaluate each complete term with that value. Do not treat the printed plus or minus sign as a claim about the numerical sign under every substitution.

How is the identity applied without losing coefficients?

Worked example 8. Expand (2n − 5m)³, where n and m are real-number variables.

Answer: Take a = 2n and b = 5m. Then (2n − 5m)³ = (2n)³ − 3(2n)²(5m) + 3(2n)(5m)² − (5m)³.

Evaluate the terms separately: (2n)³ = 8n³, 3(2n)²(5m) = 60n²m, 3(2n)(5m)² = 150nm² and (5m)³ = 125m³.

Therefore, (2n − 5m)³ = 8n³ − 60n²m + 150nm² − 125m³.

The coefficient 60 comes from 3 × 4 × 5, whereas 150 comes from 3 × 2 × 25. These calculations show why copying one middle coefficient into the other position would not preserve the identity.

For a direct check of the method, multiply (a − b) by a² − 2ab + b². Combining the resulting terms produces the same four-term expression. This route connects the difference-cube identity with the difference-square identity already established.

Note: A cube of a difference includes two mixed terms. Writing just a³ − b³ drops both of them and is not the general expansion of (a − b)³.

How can identities be chosen and expansions checked?

Which structure should be recognised first?

Before calculating, inspect the entire expression. Count the terms in a bracket, note whether it is squared or cubed, and compare separate brackets. This determines whether to use a binomial square, trinomial square, binomial cube or product identity.

A convenient numerical split should make the individual powers and products manageable. A number can be written as a sum or a difference, but the chosen form must equal the original number. The identity then preserves that value throughout the calculation.

Worked example 9. Calculate 43² using an identity.

Answer: Write 43 = 40 + 3. Then 43² = (40 + 3)² = 40² + 2 × 40 × 3 + 3² = 1600 + 240 + 9 = 1849.

Worked example 10. Calculate 199³ using an identity.

Answer: Write 199 = 200 − 1. Then 199³ = 200³ − 3(200)²(1) + 3(200)(1)² − 1³ = 8000000 − 120000 + 600 − 1 = 7880599.

Evaluate each power before combining the four signed terms. In particular, the third contribution is positive, while the final contribution is negative.

What should be checked before the final line?

1. Substitution: Identify each whole term, including its coefficient and sign, before replacing letters in the identity.
2. Powers: Apply a square or cube to the entire substituted term, not just to its variable part.
3. Cross terms: Include every required mixed product, with the correct multiplier and the sign obtained from multiplication.
4. Simplification: Combine like terms and check that the final expression contains every contribution from the preceding line.

An expansion check by ordinary distribution is especially useful when a remembered sign is uncertain. Substitution of numerical values provides another check for those values, but agreement for a single substitution does not establish an identity for all values.

Keep intermediate expressions visible. They show which identity was used and where a coefficient or sign came from. If a result disagrees with a check, return to the individual products before repeating the final addition.

Glossary

• Variable — A letter representing a number whose value may vary in an expression.
• Term — A signed part of an expression separated by addition or subtraction.
• Coefficient — The numerical factor multiplying the variable part of an algebraic term.
• Like terms — Terms with identical variable parts and powers, whose coefficients can be combined.
• Factor — An expression multiplied by another expression to form a product.
• Binomial — An algebraic expression consisting of two unlike terms with their signs.
• Trinomial — An algebraic expression consisting of three unlike terms with their signs.
• Expansion — Rewriting a product or bracketed power as a sum of terms.
• Identity — An equation that holds for all values of the variables occurring in it.
• Distributive property — The rule for multiplying each term inside a bracket by an outside factor.
• Cross term — A mixed product involving different parts of the expression being expanded.
• Cube of an expression — The product of three equal factors, each equal to that expression.

Common errors and misconceptions

• Misconception: Squaring a sum requires just the two individual squares. Correct: Include twice the product: (a + b)² = a² + 2ab + b².
• Misconception: The last term in (a − b)² is −b². Correct: The last term is +b² because (−b)(−b) = b².
• Misconception: Squaring 5x leaves its coefficient unchanged. Correct: Both factors are squared, so (5x)² = 25x².
• Misconception: The constant term in (x − 2)(x − 3) is negative. Correct: Multiplying the two negative parts gives +6; the middle term is −5x.
• Misconception: A three-term square has just three squared terms. Correct: Include three doubled pairwise products as well as the individual squares.
• Misconception: Every cross term in (a − b − c)² is negative. Correct: The product of −b and −c contributes the positive term 2bc.
• Misconception: Cubing a difference gives just a³ − b³. Correct: The full result is a³ − 3a²b + 3ab² − b³.
• Misconception: One successful numerical check proves an identity. Correct: It checks those values; distribution can establish equality for all values of the variables.

Exam-style questions with model answers

Q1. For real-number variables x and y, expand (5x + 2y)² using an identity. [2 marks]

1. Apply the sum-square identity with a = 5x and b = 2y: (5x + 2y)² = (5x)² + 2(5x)(2y) + (2y)².
2. Evaluate the individual squares and the doubled product to obtain the expansion 25x² + 20xy + 4y².

Q2. Calculate 29² using the square-of-a-difference identity, showing the substitution. [2 marks]

1. Write 29 = 30 − 1 and apply (a − b)² = a² − 2ab + b² with a = 30 and b = 1.
2. Therefore, 29² = 30² − 2 × 30 × 1 + 1² = 900 − 60 + 1 = 841.

Q3. For a real-number variable x, expand (x + 3)(x + 4). Identify the coefficient of x and the constant term. [3 marks]

1. Multiply each term of the first bracket by each term of the second bracket: (x + 3)(x + 4) = x² + 4x + 3x + 12.
2. The middle terms are like terms. Combining 4x and 3x gives 7x, so the complete expansion is x² + 7x + 12.
3. The coefficient of x is 7, obtained from 3 + 4. The constant term is 12, obtained from multiplying the two constant parts, 3 and 4.

Q4. For real-number variables p and q, expand (p + 2q)³ using the cube-of-a-sum identity. [4 marks]

1. Apply (a + b)³ = a³ + 3a²b + 3ab² + b³, substituting the whole terms a = p and b = 2q.
2. The individual cubes are p³ and (2q)³ = 8q³. Cubing the second term cubes its numerical coefficient as well as its variable.
3. The mixed terms are 3p²(2q) = 6p²q and 3p(2q)² = 12pq². They remain separate because their variable parts differ.
4. Adding these four contributions gives (p + 2q)³ = p³ + 6p²q + 12pq² + 8q³, with every term of the identity included.

Q5. For real-number variables x, y and z, expand (3x − 2y + 4z)². Show the individual squares and the three doubled products. [5 marks]

1. Use the three-term-square identity on the signed terms 3x, −2y and 4z. Its expansion requires the square of each term and twice the product of each distinct pair.
2. The individual squares are (3x)² = 9x², (−2y)² = 4y² and (4z)² = 16z². Squaring the negative coefficient in the second term gives a positive coefficient.
3. The doubled product of the first and second terms is 2(3x)(−2y) = −12xy. Retain the negative sign from the second term.
4. The other doubled products are 2(−2y)(4z) = −16yz and 2(4z)(3x) = 24xz, obtained by using the two remaining pairs.
5. Combining all six contributions gives 9x² + 4y² + 16z² − 12xy − 16yz + 24xz. The three mixed variable parts are different and remain separate.

Q6. For real-number variables n and m, expand (2n − 5m)³ using an identity, showing how both middle coefficients are obtained. [5 marks]

1. Use (a − b)³ = a³ − 3a²b + 3ab² − b³. Match the complete terms by taking a = 2n and b = 5m.
2. The first cube is (2n)³ = 8n³. The last contribution is −(5m)³ = −125m³, because the difference-cube identity subtracts the second cube.
3. The first mixed contribution is −3(2n)²(5m) = −60n²m. Its numerical coefficient comes from multiplying −3, 4 and 5 after squaring 2n.
4. The second mixed contribution is +3(2n)(5m)² = +150nm². Here the numerical coefficient comes from 3 × 2 × 25 after squaring 5m.
5. Put the four contributions together: (2n − 5m)³ = 8n³ − 60n²m + 150nm² − 125m³. The two middle terms have different powers and cannot be combined.

Key takeaways

• An identity holds for all values of its variables; distribution explains why an expansion is generally valid.
• In a binomial square, retain both individual squares and twice the product, with the appropriate sign.
• Substitute complete terms into an identity, keeping each numerical coefficient and sign attached to its variable part.
• For products with a common first term, use the signed sum and signed product of the other terms.
• A trinomial square contains three individual squares and three doubled pairwise products; check each pair separately.
• A binomial cube includes two mixed terms as well as the individual cubes of its terms.
• Numerical calculations become easier when a number is expressed as a convenient sum or difference before expansion.
• Combine terms only when their variable parts match exactly, including every power appearing in those parts.

Test yourself

What is the difference between an equation and an identity?

An identity is an equation true for all values of its variables. An equation need not hold for every value.

Why does (a + b)² include 2ab?

Distribution produces the two equal products ab and ba. Combining them gives the middle term 2ab.

Why is the last term in (a − b)² positive?

It comes from (−b)(−b), which equals b² because the two negative factors give a positive product.

What is the expansion of (x − 2)(x − 3)?

It is x² − 5x + 6: combine −3x and −2x, and multiply −2 by −3.

Which doubled products belong in (a + b + c)²?

The required doubled products are 2ab, 2bc and 2ca, one for each distinct pair of terms.

What are the cross-term signs in (a − b − c)²?

The cross terms are −2ab, +2bc and −2ca. The positive one comes from multiplying the two negative terms.

Why does (a − b)³ contain +3ab²?

Replacing b by −b gives 3a(−b)². Squaring the negative term produces b², leaving the displayed plus sign.

Can a single numerical substitution prove an identity?

No. It checks equality for those values. An algebraic derivation using distribution establishes the identity for all values.