MathsICSE Class 8 · Chapter 11

Algebraic Expressions and Identities ICSE Class 8: Multiplication, Division and Identities

Updated 15 October 2026 · Fact-checked

An algebraic identity is an equation that is true for every value of its variables. ICSE Class 8 uses (a + b)² = a² + 2ab + b², (a − b)² = a² − 2ab + b² and a² − b² = (a − b)(a + b), with rules for multiplying and dividing expressions, to simplify and evaluate quickly.

What Algebraic Expressions and Identities is about

An algebraic expression is built from numbers and letters (variables) using $+$ , $-$ , $\times$ and $\div$ , for example $3 x^{2} - 5 x y + 7$ . Its parts separated by $+$ or $-$ signs are terms; the number in front of a term is its coefficient. Terms with the same letters raised to the same powers are like terms ( $4 x^{2} y$ and $- x^{2} y$ ) and can be added; unlike terms cannot.

The CISCE Class VIII curriculum asks you to:

multiply algebraic expressions: monomial × monomial, monomial × polynomial, and polynomial × polynomial;
divide algebraic expressions, with integer coefficients;
use the identities $(a \pm b)^{2} = a^{2} \pm 2 ab + b^{2}$ and $a^{2} - b^{2} = (a - b) (a + b)$ .

An identity is true for all values of the letters, unlike an equation such as $2 x + 1 = 7$ , which is true only for $x = 3$ . Identities let you expand squares in one line, work out things like $99 8^{2}$ or $47 \times 53$ in your head, and find $a^{2} + b^{2}$ from $a + b$ and $ab$ . They lead straight into the next chapter, Factorisation.

Weightage: CISCE publishes no question paper, marking scheme or chapter weightage for Class VIII Mathematics: your school sets and marks the exam. Algebraic Expressions and Identities sits in the Algebra theme of the CISCE Curriculum for Upper Primary Classes (VI-VIII).Source: CISCE Curriculum for Upper Primary Classes (VI-VIII), RDCD 2016

Key concepts

Multiplying monomials

To multiply monomials:
1. Multiply the coefficients (watch the signs).
2. Multiply the variables, adding the powers of the same letter: $x^{2} \times x = x^{3}$ .

Sign rules: $(+) (+) = +$ , $(-) (-) = +$ , $(+) (-) = -$ .

The order of the factors does not matter, so you can group numbers with numbers and letters with letters.

Worked example

Multiply $- 3 x^{2} y$ by $4 x y^{3}$ .

Try it first, then show the solutionSolution

Coefficients: $(- 3) \times 4 = - 12$ .
Variables: $x^{2} \times x = x^{3}$ and $y \times y^{3} = y^{4}$ .
Product $= - 12 x^{3} y^{4}$ .

Answer: $- 12 x^{3} y^{4}$

Multiplying a polynomial by a monomial

Use the distributive law: multiply the monomial by every term inside the bracket.
$a (b + c - d) = ab + a c - a d$ .

Keep track of the sign of each term. A minus sign in front of the monomial changes the sign of every term.

Worked example

Find $2 a (3 a^{2} - 4 ab + 5)$ .

Try it first, then show the solutionSolution

$2 a \times 3 a^{2} = 6 a^{3}$ .
$2 a \times (- 4 ab) = - 8 a^{2} b$ .
$2 a \times 5 = 10 a$ . So the product is $6 a^{3} - 8 a^{2} b + 10 a$ .

Answer: $6 a^{3} - 8 a^{2} b + 10 a$

Multiplying two polynomials

Multiply each term of the first bracket by each term of the second, then collect like terms.

For two binomials: $(a + b) (c + d) = a c + a d + b c + b d$ .

A useful pattern: $(x + a) (x + b) = x^{2} + (a + b) x + ab$ .

Check: the number of products before collecting is (terms in first) × (terms in second). Putting in a simple value such as $x = 1$ on both sides is a quick check.

Worked example

Multiply $(2 x + 3) (x - 4)$ .

Try it first, then show the solutionSolution

$2 x \times x = 2 x^{2}$ ; $2 x \times (- 4) = - 8 x$ .
$3 \times x = 3 x$ ; $3 \times (- 4) = - 12$ .
Collect: $2 x^{2} - 8 x + 3 x - 12 = 2 x^{2} - 5 x - 12$ .
Check with $x = 1$ : $(5) (- 3) = - 15$ and $2 - 5 - 12 = - 15$ ✓.

Answer: $2 x^{2} - 5 x - 12$

Dividing algebraic expressions

Monomial ÷ monomial: divide coefficients, subtract powers: $\frac{12 x ^{3} y ^{2}}{4 x y} = 3 x^{2} y$ .

Polynomial ÷ monomial: divide each term by the monomial.

Polynomial ÷ polynomial: use long division.
1. Arrange both in descending powers of $x$ (leave a gap or write $0 x$ for a missing power).
2. Divide the first term of the dividend by the first term of the divisor; write it in the quotient.
3. Multiply the divisor by it and subtract.
4. Bring down the next term and repeat until the remainder is 0 or of lower degree than the divisor.

Check: divisor × quotient + remainder = dividend.

Worked example

Divide $x^{2} + 5 x + 6$ by $x + 2$ .

Try it first, then show the solutionSolution

$x^{2} \div x = x$ . Multiply: $x (x + 2) = x^{2} + 2 x$ . Subtract: $3 x + 6$ .
$3 x \div x = 3$ . Multiply: $3 (x + 2) = 3 x + 6$ . Subtract: 0.
Quotient $x + 3$ , remainder 0. Check: $(x + 2) (x + 3) = x^{2} + 5 x + 6$ ✓.

Answer: $x + 3$

The identities (a + b)² and (a − b)²

$(a + b)^{2} = (a + b) (a + b) = a^{2} + 2 ab + b^{2}$
$(a - b)^{2} = a^{2} - 2 ab + b^{2}$

In words: square the first, twice the product, square the second. The middle term $2 ab$ is the one students forget.

They work with any terms in place of $a$ and $b$ : $(3 x + 2 y)^{2}$ has $a = 3 x$ , $b = 2 y$ .

They also square numbers quickly: $10 2^{2} = (100 + 2)^{2} = 10000 + 400 + 4 = 10404$ .

Worked example

Expand $(3 x + 2 y)^{2}$ .

Try it first, then show the solutionSolution

$a = 3 x$ , $b = 2 y$ .
$a^{2} = 9 x^{2}$ , $2 ab = 2 (3 x) (2 y) = 12 x y$ , $b^{2} = 4 y^{2}$ .
$(3 x + 2 y)^{2} = 9 x^{2} + 12 x y + 4 y^{2}$ .

Answer: $9 x^{2} + 12 x y + 4 y^{2}$

The identity a² − b² = (a − b)(a + b) and its uses

$(a - b) (a + b) = a^{2} + ab - ab - b^{2} = a^{2} - b^{2}$ . The middle terms cancel.

Uses:

Products of numbers equally spaced around a round number: $47 \times 53 = (50 - 3) (50 + 3) = 2500 - 9 = 2491$ .
Differences of squares: $10 3^{2} - 9 7^{2} = (103 - 97) (103 + 97) = 6 \times 200 = 1200$ .

Combining identities gives values without knowing $a$ and $b$ :

$a^{2} + b^{2} = (a + b)^{2} - 2 ab = (a - b)^{2} + 2 ab$
$(a + b)^{2} - (a - b)^{2} = 4 ab$

Worked example

If $a + b = 7$ and $ab = 10$ , find $a^{2} + b^{2}$ .

Try it first, then show the solutionSolution

$a^{2} + b^{2} = (a + b)^{2} - 2 ab$ .
$= 49 - 20 = 29$ .
Check with $a = 5$ , $b = 2$ : $25 + 4 = 29$ ✓.

Answer: 29

Maths in real life: Algebraic Expressions and Identities

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

(a + b)² with numbers

Suppose you need the area of a square tile of side 31 cm without a calculator.

Find the area.

Try it first, then show the solutionSolution

$3 1^{2} = (30 + 1)^{2} = 900 + 60 + 1$ .
$= 961$ cm².

Answer: 961 cm²

Memory hook: Split into a round number plus a little.

a² − b²

Imagine a photo print that is 25 cm long and 15 cm wide.

Find its area quickly using an identity.

Try it first, then show the solutionSolution

$25 \times 15 = (20 + 5) (20 - 5)$ .
$= 2 0^{2} - 5^{2} = 400 - 25 = 375$ cm².

Answer: 375 cm²

Memory hook: Numbers equally either side of a round number: square of the middle minus square of the gap.

Expanding a product

Suppose a square garden has side $x$ m, and you make each side 2 m longer.

By how much does the area grow? Find the increase when $x = 10$ .

Try it first, then show the solutionSolution

New area $= (x + 2)^{2} = x^{2} + 4 x + 4$ ; increase $= 4 x + 4$ .
When $x = 10$ : $40 + 4 = 44$ m² (check: $144 - 100 = 44$ ).

Answer: $(4 x + 4)$ m²; 44 m² when $x = 10$

Memory hook: The extra area is the 2ab strips plus the small b² corner.

Topics in this chapter

Algebraic Expressions and Identities formula sheet

Square of a sum: $(a + b)^{2} = a^{2} + 2 ab + b^{2}$; Do not forget 2ab.
Square of a difference: $(a - b)^{2} = a^{2} - 2 ab + b^{2}$; Last term is +b².
Difference of squares: $a^{2} - b^{2} = (a - b) (a + b)$; Middle terms cancel.
Product of two binomials: $(x + a) (x + b) = x^{2} + (a + b) x + ab$; Add for the middle, multiply for the end.
Sum of squares: $a^{2} + b^{2} = (a + b)^{2} - 2 ab$; Also (a − b)² + 2ab.
Difference of the two squares: $(a + b)^{2} - (a - b)^{2} = 4 ab$; Finds ab from a + b and a − b.

How the school exam asks Algebraic Expressions and Identities

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Multiply expressions	Multiply (3x − 2)(2x² + x − 4) and simplify.	Every term by every term, then collect like terms; check with x = 1.
Divide expressions	Divide 15a³b² − 10a²b³ by 5ab, or 6x³ + 5x² − x − 1 by 2x + 1.	Term by term for a monomial divisor; long division for a binomial.
Expand using identities	Expand (2a − 5b)² or (x + 4)(x − 7).	Identify a and b, then apply the identity in one line.
Evaluate numbers	Find 998², 105 × 95 or 103² − 97² using identities.	Write as (round ± small) and apply an identity.
Find values	If x + 1/x = 5, find x² + 1/x²; if a − b = 4 and ab = 21, find a² + b².	Square the given expression and rearrange.

Tricks to remember Algebraic Expressions and Identities

Short memory aids. Say each one aloud once, then use it on a question the same day.

First², twice the product, last²
Say it aloud for every (a ± b)²: 'first squared, twice first times last, last squared'.
Use it when: Never losing the middle term.
Plus-minus pair kills the middle
(a + b)(a − b): the cross terms +ab and −ab cancel, leaving a² − b².
Use it when: Spotting the difference of squares.
Divide, multiply, subtract, bring down
Long division of polynomials repeats the same four moves as number long division.
Use it when: Dividing by a binomial.
Check with x = 1
Put x = 1 (or any small number) into both sides; they must give the same value.
Use it when: Catching slips in products and quotients.

Algebraic Expressions and Identities: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Product of monomials · 1 mark · MCQExam level

$(- 2 a^{2} b) \times (3 a b^{2})$ equals:
(a) $6 a^{3} b^{3}$
(b) $- 6 a^{3} b^{3}$
(c) $- 6 a^{2} b^{2}$
(d) $- 5 a^{3} b^{3}$

Try it first, then show the solutionSolution

Coefficients: $(- 2) (3) = - 6$ .
Variables: $a^{2} \times a = a^{3}$ , $b \times b^{2} = b^{3}$ .

Answer: (b)

Example 2: Binomial × trinomial · 3 marks · Short answerExam level

Multiply $(3 x - 2) (2 x^{2} + x - 4)$ .

Try it first, then show the solutionSolution

$3 x (2 x^{2} + x - 4) = 6 x^{3} + 3 x^{2} - 12 x$ .
$- 2 (2 x^{2} + x - 4) = - 4 x^{2} - 2 x + 8$ .
Add: $6 x^{3} - x^{2} - 14 x + 8$ .

Answer: $6 x^{3} - x^{2} - 14 x + 8$

Example 3: Square of a difference · 3 marks · Short answerExam level

Using an identity, expand $(2 a - 5 b)^{2}$ .

Try it first, then show the solutionSolution

Use $(a - b)^{2} = a^{2} - 2 ab + b^{2}$ with first term $2 a$ and second term $5 b$ .
$(2 a)^{2} = 4 a^{2}$ ; $2 (2 a) (5 b) = 20 ab$ ; $(5 b)^{2} = 25 b^{2}$ .
$(2 a - 5 b)^{2} = 4 a^{2} - 20 ab + 25 b^{2}$ .

Answer: $4 a^{2} - 20 ab + 25 b^{2}$

Example 4: Square a big number · 3 marks · Short answerHarder exam level

Using a suitable identity, find $99 8^{2}$ .

Try it first, then show the solutionSolution

$99 8^{2} = (1000 - 2)^{2}$ .
$= 1000000 - 2 \times 1000 \times 2 + 4 = 1000000 - 4000 + 4$ .
$= 996004$ .

Answer: 996004

Example 5: Long division · 4 marks · Long answerHarder exam level

Divide $6 x^{3} + 5 x^{2} - x - 1$ by $2 x + 1$ and check your answer.

Try it first, then show the solutionSolution

$6 x^{3} \div 2 x = 3 x^{2}$ ; $3 x^{2} (2 x + 1) = 6 x^{3} + 3 x^{2}$ ; subtract: $2 x^{2} - x$ .
$2 x^{2} \div 2 x = x$ ; $x (2 x + 1) = 2 x^{2} + x$ ; subtract: $- 2 x - 1$ .
$- 2 x \div 2 x = - 1$ ; $- 1 (2 x + 1) = - 2 x - 1$ ; subtract: 0.
Quotient $3 x^{2} + x - 1$ . Check: $(2 x + 1) (3 x^{2} + x - 1) = 6 x^{3} + 2 x^{2} - 2 x + 3 x^{2} + x - 1 = 6 x^{3} + 5 x^{2} - x - 1$ ✓.

Answer: $3 x^{2} + x - 1$ , remainder 0

Example 6: x + 1/x · 4 marks · Long answerHarder exam level

If $x + \frac{1}{x} = 5$ , find (i) $x^{2} + \frac{1}{x ^{2}}$ (ii) $(x - \frac{1}{x})^{2}$ .

Try it first, then show the solutionSolution

Square: $(x + \frac{1}{x})^{2} = x^{2} + 2 + \frac{1}{x ^{2}}$ , because $2 \times x \times \frac{1}{x} = 2$ .
(i) $25 = x^{2} + \frac{1}{x ^{2}} + 2$ , so $x^{2} + \frac{1}{x ^{2}} = 23$ .
(ii) $(x - \frac{1}{x})^{2} = x^{2} + \frac{1}{x ^{2}} - 2$ .
$= 23 - 2 = 21$ .

Answer: (i) 23 (ii) 21

Example 7: From a + b and a − b · 5 marks · Long answerHOTS / competency

Show that $(a + b)^{2} - (a - b)^{2} = 4 ab$ . Hence, if $a + b = 9$ and $a - b = 3$ , find $ab$ and $a^{2} + b^{2}$ .

Try it first, then show the solutionSolution

$(a + b)^{2} - (a - b)^{2} = (a^{2} + 2 ab + b^{2}) - (a^{2} - 2 ab + b^{2}) = 4 ab$ .
$4 ab = 81 - 9 = 72$ , so $ab = 18$ .
$a^{2} + b^{2} = (a + b)^{2} - 2 ab = 81 - 36$ .
$= 45$ .
Check: $a = 6$ , $b = 3$ gives $ab = 18$ and $36 + 9 = 45$ ✓.

Answer: $ab = 18$ ; $a^{2} + b^{2} = 45$

Case-study practice (Section B style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 10-mark question split into 3 + 3 + 4 marks.

Planning a garden

Suppose a square flower bed has side $(2 x + 3)$ m, and a rectangular vegetable bed next to it measures $(2 x + 3)$ m by $(2 x - 3)$ m, where $x > 2$ .

(i) · 3 marks
Find the area of the square bed as an expression.
Answer
Answer: $(4 x^{2} + 12 x + 9)$ m²
$(2 x + 3)^{2} = 4 x^{2} + 2 (2 x) (3) + 9 = 4 x^{2} + 12 x + 9$ .
(ii) · 3 marks
Find the area of the rectangular bed and how much larger the square bed is.
Answer
Answer: $(4 x^{2} - 9)$ m²; larger by $(12 x + 18)$ m²
$(2 x + 3) (2 x - 3) = 4 x^{2} - 9$ . Difference $= (4 x^{2} + 12 x + 9) - (4 x^{2} - 9) = 12 x + 18$ .
(iii) · 4 marks
Find both areas when $x = 5$ and check the difference with your expression.
Answer
Answer: 169 m² and 91 m²; difference 78 m²
Sides 13 m and 13 m × 7 m. Areas $169$ and $91$ m². Difference $= 78$ , and $12 (5) + 18 = 78$ ✓.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Simplify $(x + 2) (x - 3) - (x - 1) (x + 4)$ . (3 marks)

Step	Marks
$(x + 2) (x - 3) = x^{2} - x - 6$	1
$(x - 1) (x + 4) = x^{2} + 3 x - 4$	1
Difference $= x^{2} - x - 6 - x^{2} - 3 x + 4 = - 4 x - 2$	1

Using identities, find $10 3^{2}$ and $9 7^{2}$ . Then find $10 3^{2} - 9 7^{2}$ using a different identity and check that the answers agree. (4 marks)

Step	Marks
$10 3^{2} = (100 + 3)^{2} = 10000 + 600 + 9 = 10609$	1
$9 7^{2} = (100 - 3)^{2} = 10000 - 600 + 9 = 9409$	1
$10 3^{2} - 9 7^{2} = (103 - 97) (103 + 97) = 6 \times 200 = 1200$	1
Check: $10609 - 9409 = 1200$ ✓	1

Divide $6 x^{2} - x - 15$ by $2 x + 3$ and verify the result. (5 marks)

Step	Marks
$6 x^{2} \div 2 x = 3 x$ ; $3 x (2 x + 3) = 6 x^{2} + 9 x$	1
Subtract: $- x - 9 x = - 10 x$ ; bring down $- 15$ : $- 10 x - 15$	1
$- 10 x \div 2 x = - 5$ ; $- 5 (2 x + 3) = - 10 x - 15$ ; remainder 0	1
Quotient $= 3 x - 5$	1
Verify: $(2 x + 3) (3 x - 5) = 6 x^{2} - 10 x + 9 x - 15 = 6 x^{2} - x - 15$ ✓	1

Where students lose marks in Algebraic Expressions and Identities

The slips examiners see most often, and what to write instead.

Mistake: Writing $(a + b)^{2} = a^{2} + b^{2}$
The middle term $2 ab$ is missing.
Wrong working
$(x + 3)^{2} = x^{2} + 9$
Right working
$(x + 3)^{2} = x^{2} + 6 x + 9$
Do this: Say 'twice the product' every time.
Usually costs: The whole expansion.
Mistake: Wrong sign on the last term of $(a - b)^{2}$
$(- b)^{2} = + b^{2}$ .
Wrong working
$(2 a - 5 b)^{2} = 4 a^{2} - 20 ab - 25 b^{2}$
Right working
$4 a^{2} - 20 ab + 25 b^{2}$
Do this: Only the middle term is negative.
Usually costs: 1 mark.
Mistake: Not multiplying every term
In $2 a (3 a^{2} - 4 ab + 5)$ the last term is often left as $5$ .
Do this: Draw an arrow from the monomial to each term inside the bracket.
Usually costs: 1 mark per missing term.
Mistake: Subtracting wrongly in long division
Subtracting a whole line means changing every sign of that line.
Wrong working
$(6 x^{2} - x) - (6 x^{2} + 9 x) = 8 x$
Right working
$= - 10 x$
Do this: Change the signs of the line being subtracted, then add.
Usually costs: The quotient.
Mistake: Adding unlike terms
$3 x^{2}$ and $3 x$ are unlike terms; they cannot be combined.
Do this: Collect only terms with exactly the same letters and powers.
Usually costs: 1 mark.

Practice questions on Algebraic Expressions and Identities

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
$(x + 3) (x - 3)$ equals:
(a) $x^{2} + 9$
(b) $x^{2} - 9$
(c) $x^{2} - 6 x + 9$
(d) $x^{2} - 6$
Hint
$a^{2} - b^{2}$ .
Answer
(b)
Q2 · 2 marks · Short answerExam level
Multiply $4 p^{2} q$ by $(p - 2 q + 3)$ .
Hint
Multiply each term.
Answer
$4 p^{3} q - 8 p^{2} q^{2} + 12 p^{2} q$ .
Q3 · 2 marks · Short answerExam level
Divide $15 a^{3} b^{2} - 10 a^{2} b^{3}$ by $5 ab$ .
Hint
Divide each term.
Answer
$3 a^{2} b - 2 a b^{2}$ .
Q4 · 2 marks · Short answerHarder exam level
Find $(x + 4) (x - 7)$ using $(x + a) (x + b) = x^{2} + (a + b) x + ab$ .
Hint
$a = 4$ , $b = - 7$ .
Answer
$x^{2} - 3 x - 28$ .
Q5 · 1 mark · Assertion-reasonHarder exam level
Assertion (A): $(a - b)^{2} = a^{2} - b^{2}$ .
Reason (R): $(a - b)^{2} = a^{2} - 2 ab + b^{2}$ .
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
Try a = 3, b = 1.
Answer
(b). $(3 - 1)^{2} = 4$ but $9 - 1 = 8$ , so (A) is false; (R) is the correct identity.
Q6 · 2 marks · Short answerHarder exam level
Using a suitable identity, find $105 \times 95$ .
Hint
$(100 + 5) (100 - 5)$ .
Answer
$10000 - 25 = 9975$ .
Q7 · 3 marks · Short answerHarder exam level
If $a - b = 4$ and $ab = 21$ , find $a^{2} + b^{2}$ .
Hint
$(a - b)^{2} + 2 ab$ .
Answer
$16 + 42 = 58$ .
Q8 · 3 marks · Short answerHarder exam level
Simplify $(2 x + 3 y)^{2} - (2 x - 3 y)^{2}$ .
Hint
The squared terms cancel.
Answer
$(4 x^{2} + 12 x y + 9 y^{2}) - (4 x^{2} - 12 x y + 9 y^{2}) = 24 x y$ .
Q9 · 4 marks · Long answerHarder exam level
Divide $x^{3} - 8$ by $x - 2$ .
Hint
Write the dividend as $x^{3} + 0 x^{2} + 0 x - 8$ .
Answer
Quotient $x^{2} + 2 x + 4$ , remainder 0. Check: $(x - 2) (x^{2} + 2 x + 4) = x^{3} - 8$ .
Q10 · 2 marks · Short answerHarder exam level
Find the value of $4 x^{2} + 12 x + 9$ when $x = 2.5$ , using an identity.
Hint
It is a perfect square.
Answer
$(2 x + 3)^{2} = (5 + 3)^{2} = 64$ .
Q11 · 4 marks · Long answerHOTS / competency
If $x - \frac{1}{x} = 3$ , find (i) $x^{2} + \frac{1}{x ^{2}}$ (ii) $x^{4} + \frac{1}{x ^{4}}$ .
Hint
Square twice.
Answer
(i) $9 = x^{2} + \frac{1}{x ^{2}} - 2$ , so 11. (ii) $121 = x^{4} + \frac{1}{x ^{4}} + 2$ , so 119.
Q12 · 3 marks · Short answerHOTS / competency
Prove that $(a + b) (a - b) + (b + c) (b - c) + (c + a) (c - a) = 0$ .
Hint
Use $a^{2} - b^{2}$ three times.
Answer
$(a^{2} - b^{2}) + (b^{2} - c^{2}) + (c^{2} - a^{2}) = 0$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Expand (a + b)².Show answer
a² + 2ab + b².
2.Expand (a − b)².Show answer
a² − 2ab + b².
3.Factorise a² − b² as a product.Show answer
(a − b)(a + b).
4.What is 5x²y × 3xy?Show answer
15x³y².
5.Divide 18a³b² by 6ab.Show answer
3a²b.
6.Find 101².Show answer
10201.
7.Find 52 × 48.Show answer
2496 (50² − 2²).
8.Are 4x²y and −7yx² like terms?Show answer
Yes.

Best order to study it

Revise terms, coefficients and like terms.
Multiply monomials, then monomial × polynomial, then polynomial × polynomial.
Divide by a monomial, then learn long division by a binomial.
Learn (a ± b)² and a² − b² and apply them to algebra and to numbers.
Finish with finding values such as a² + b² and x² + 1/x².

Quick revision checklist

The night before the exam: can you do each of these without looking?

Did you multiply every term by every term?
Did you add powers of the same letter when multiplying?
Did you include the 2ab term in (a ± b)²?
Did you change all the signs when subtracting in long division?
Did you check a product or quotient by putting x = 1?
Can you use a² − b² to multiply numbers like 47 × 53?

Algebraic Expressions and Identities: frequently asked questions

What is the difference between an identity and an equation?

An identity is true for every value of the variables, like (a + b)² = a² + 2ab + b². An equation is true only for particular values, like 2x + 1 = 7, which holds only for x = 3.

Why are the coefficients integers in Class 8?

The CISCE Class VIII syllabus limits multiplication and division to integer coefficients so that the focus stays on the method, not on fraction arithmetic.

How can I check a long division answer?

Multiply the divisor by the quotient and add the remainder. You must get the dividend back.

Is (x + a)(x + b) an identity too?

Yes: (x + a)(x + b) = x² + (a + b)x + ab for all x. It is just the product of two binomials, and it is used again in factorisation.

Why learn identities if I can multiply out?

They are faster and less error-prone, and the same patterns are needed backwards in factorisation.

Based on: CISCE Curriculum for Upper Primary Classes (VI-VIII), Mathematics (Class VIII); CISCE RDCD: Curriculum Pre-School to Class VIII