Important Questions Class 9 Maths Chapter 4 Exploring Algebraic Identities 2026-2027

Exploring Algebraic Identities teaches students how algebraic rules stay true for all values of variables. These identities help expand expressions, factorise polynomials, simplify rational expressions and calculate squares or cubes quickly.

Important questions class 9 maths chapter 4 help students practise identity-based expansion, factorisation, algebra tiles, products, rational expression simplification and application questions. This chapter is important because algebraic identities support polynomials, quadratic equations, mensuration and coordinate geometry.

Class 9 Maths Chapter 4 begins with patterns in consecutive square numbers. Students see that algebra can explain why number patterns always work.

The chapter then uses square and rectangle models to visualise identities. Students learn why ((a + b)^2 = a^2 + 2ab + b^2), how ((a - b)^2) works, and how to expand ((a + b + c)^2). Later sections introduce algebra tiles, cubic identities and rational algebraic expressions.

Key Takeaways from Class 9 Maths Chapter 4

Important Questions Class 9 Maths Chapter 4 with Answers

These questions cover the main ideas of the chapter. Students should write the identity first, then substitute values carefully.

Important Questions Class 9 Maths Chapter 4: Basic Concepts

Q1. What is an algebraic identity?

An algebraic identity is an equation that remains true for all values of the variables in it.

For example:

(x + y)^2 = x^2 + 2xy + y^2

This is true for every value of (x) and (y). So, it is an identity.

Q2. What is the difference between an equation and an identity?

An equation is true only for some values of the variable. An identity is true for all values.

For example:

x^2 - 1 = 24

This is true only for (x = 5) or (x = -5).

But:

(a + b)^2 = a^2 + 2ab + b^2

This is true for all values of (a) and (b).

Q3. Write the identity for ((a + b)^2).

(a + b)^2 = a^2 + 2ab + b^2

This identity helps expand the square of a binomial.

Example:

(5x + 2y)^2 = 25x^2 + 20xy + 4y^2

Q4. Write the identity for ((a - b)^2).

(a - b)^2 = a^2 - 2ab + b^2

This identity helps calculate squares close to round numbers.

Example:

29^2 = (30 - 1)^2

= 900 - 60 + 1 = 841

Q5. Write the identity for ((a + b + c)^2).

(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

This identity helps expand the square of three terms.

Algebraic Identities Class 9 Important Questions

Algebraic identities class 9 questions test formula use. Always identify (a), (b) and (c) before expanding.

Q1. Expand ((7x + 4y)^2).

Use:

(a + b)^2 = a^2 + 2ab + b^2

Here, (a = 7x) and (b = 4y).

(7x + 4y)^2 = (7x)^2 + 2(7x)(4y) + (4y)^2

= 49x^2 + 56xy + 16y^2

Q2. Expand ((3x - 2y)^2).

Use:

(a - b)^2 = a^2 - 2ab + b^2

Here, (a = 3x) and (b = 2y).

(3x - 2y)^2 = (3x)^2 - 2(3x)(2y) + (2y)^2

= 9x^2 - 12xy + 4y^2

Q3. Expand ((p + 3q + 7r)^2).

Use:

(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

Here, (a = p), (b = 3q), and (c = 7r).

(p + 3q + 7r)^2

= p^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14pr

Q4. Expand ((3x - 2y + 4z)^2).

Here, (a = 3x), (b = -2y), and (c = 4z).

(3x - 2y + 4z)^2

= 9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24xz

Q5. Find (64^2) using an identity.

64^2 = (60 + 4)^2

= 60^2 + 2(60)(4) + 4^2

= 3600 + 480 + 16

= 4096

Class 9 Maths Chapter 4 Question Answer on Factorisation

Factorisation means writing an expression as a product of factors. Identities make factorisation faster and cleaner.

Factorisation Class 9 Maths Questions Using Identities

Q1. Factorise (x^2 + 4x + 4).

Compare with:

a^2 + 2ab + b^2 = (a + b)^2

x^2 + 4x + 4 = x^2 + 2(x)(2) + 2^2

= (x + 2)^2

Q2. Factorise (36x^2 + 12x + 1).

36x^2 + 12x + 1 = (6x)^2 + 2(6x)(1) + 1^2

= (6x + 1)^2

Q3. Factorise (9x^2 + 24xy + 16y^2).

9x^2 + 24xy + 16y^2 = (3x)^2 + 2(3x)(4y) + (4y)^2

= (3x + 4y)^2

Q4. Factorise (16y^2 - 24y + 9).

16y^2 - 24y + 9 = (4y)^2 - 2(4y)(3) + 3^2

= (4y - 3)^2

Q5. Factorise (49g^2 + 14gh + h^2).

49g^2 + 14gh + h^2 = (7g)^2 + 2(7g)(h) + h^2

= (7g + h)^2

Factorisation Class 9 Maths Questions Using Middle Term Splitting

Middle-term splitting helps factorise quadratic expressions. Find two numbers whose sum is the coefficient of (x) and product is the constant term.

Q1. Factorise (x^2 + 7x + 12).

We need two numbers whose sum is 7 and product is 12.

The numbers are 3 and 4.

x^2 + 7x + 12 = x^2 + 3x + 4x + 12

= x(x + 3) + 4(x + 3)

= (x + 3)(x + 4)

Q2. Factorise (x^2 + 11x + 30).

We need two numbers whose sum is 11 and product is 30.

The numbers are 5 and 6.

x^2 + 11x + 30 = x^2 + 5x + 6x + 30

= x(x + 5) + 6(x + 5)

= (x + 5)(x + 6)

Q3. Factorise (x^2 - 5x + 6).

We need two numbers whose sum is (-5) and product is 6.

The numbers are (-2) and (-3).

x^2 - 5x + 6 = x^2 - 2x - 3x + 6

= x(x - 2) - 3(x - 2)

= (x - 2)(x - 3)

Q4. Factorise (r^2 - r - 42).

We need two numbers whose sum is (-1) and product is (-42).

The numbers are 6 and (-7).

r^2 - r - 42 = r^2 + 6r - 7r - 42

= r(r + 6) - 7(r + 6)

= (r + 6)(r - 7)

Q5. Factorise (6x^2 + 7x + 2).

Multiply the coefficient of (x^2) and the constant term:

6 × 2 = 12

We need two numbers whose sum is 7 and product is 12.

The numbers are 3 and 4.

6x^2 + 7x + 2 = 6x^2 + 3x + 4x + 2

= 3x(2x + 1) + 2(2x + 1)

= (3x + 2)(2x + 1)

Algebra Tiles Class 9 Questions

Algebra tiles class 9 questions help students see algebra as area and arrangement. This visual method is useful for understanding expansion and factorisation.

Q1. How do algebra tiles show ((x + 2)(x + 3))?

Use one (x^2) tile, five (x) tiles and six unit tiles.

The rectangle has sides (x + 2) and (x + 3). Its area is:

x^2 + 5x + 6

So:

(x + 2)(x + 3) = x^2 + 5x + 6

Q2. How do algebra tiles help factorise (x^2 + 7x + 12)?

Arrange one (x^2) tile, seven (x) tiles and twelve unit tiles into a rectangle.

The rectangle can have side lengths (x + 3) and (x + 4).

So:

x^2 + 7x + 12 = (x + 3)(x + 4)

Q3. Why are algebra tiles useful in this chapter?

Algebra tiles help students connect expressions with area.

They make factorisation easier because students can see how terms form a rectangle. This reduces blind formula use.

More Identities Class 9 Maths Important Questions

This section covers cubic identities and identities involving three variables. These questions often appear as higher-order problems.

Q1. Write the identity for (x^3 - y^3).

x^3 - y^3 = (x - y)(x^2 + xy + y^2)

This identity helps factor the difference of cubes.

Q2. Write the identity for (x^3 + y^3).

x^3 + y^3 = (x + y)(x^2 - xy + y^2)

This identity helps factor the sum of cubes.

Q3. Write the identity for ((x + y)^3).

(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3

This identity comes from multiplying ((x + y)) with ((x + y)^2).

Q4. Write the identity for ((x - y)^3).

(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3

The signs appear alternately.

Q5. Factorise (27b^3 - (1)/(64b^3)).

Write:

27b^3 = (3b)^3

(1)/(64b^3) = ((1)/(4b))^3

Use:

x^3 - y^3 = (x - y)(x^2 + xy + y^2)

27b^3 - (1)/(64b^3)

= (3b - (1)/(4b))(9b^2 + (3)/(4) + (1)/(16b^2))

Class 9 Maths Chapter 4 MCQs with Answers

Class 9 Maths Chapter 4 MCQs test identity recall, expansion and factorisation. These are useful for quick revision before long questions.

Q1. Which of the following is an identity?

(a) (x + 2 = 5)

(b) (x^2 - 1 = 0)

(c) ((a + b)^2 = a^2 + 2ab + b^2)

(d) (x = 7)

Answer: (c)

It is true for all values of (a) and (b).

Q2. The expansion of ((x - 5)^2) is:

(a) (x^2 - 25)

(b) (x^2 + 10x + 25)

(c) (x^2 - 10x + 25)

(d) (x^2 + 25)

Answer: (c)

(x - 5)^2 = x^2 - 10x + 25

Q3. The factors of (x^2 + 9x + 20) are:

(a) ((x + 2)(x + 10))

(b) ((x + 4)(x + 5))

(c) ((x - 4)(x - 5))

(d) ((x + 1)(x + 20))

Answer: (b)

4 + 5 = 9 and 4 × 5 = 20.

Q4. The value of (105^2) is:

(a) 10025

(b) 11025

(c) 12025

(d) 10525

Answer: (b)

105^2 = (100 + 5)^2 = 11025

Q5. The identity (x^3 + y^3) equals:

(a) ((x - y)(x^2 + xy + y^2))

(b) ((x + y)(x^2 - xy + y^2))

(c) ((x + y)(x^2 + xy + y^2))

(d) ((x - y)(x^2 - xy + y^2))

Answer: (b)

x^3 + y^3 = (x + y)(x^2 - xy + y^2)

Q6. Which identity is used for (104 × 96)?

(a) ((a + b)^2)

(b) ((a - b)^2)

(c) ((a + b)(a - b))

(d) ((a + b + c)^2)

Answer: (c)

104 × 96 = (100 + 4)(100 - 4)

So, use (a^2 - b^2).

Class 9 Maths Chapter 4 Extra Questions with Answers

Class 9 Maths Chapter 4 extra questions help students practise mental calculation, expansion, factorisation and products using identities.

Q1. Find (79^2) using an identity.

79^2 = (80 - 1)^2

= 80^2 - 2(80)(1) + 1^2

= 6400 - 160 + 1

= 6241

Q2. Find (117^2) using an identity.

117^2 = (100 + 10 + 7)^2

= 100^2 + 10^2 + 7^2 + 2(100)(10) + 2(10)(7) + 2(7)(100)

= 10000 + 100 + 49 + 2000 + 140 + 1400

= 13689

Q3. Find (104 × 96) using a suitable identity.

104 × 96 = (100 + 4)(100 - 4)

Use:

(a + b)(a - b) = a^2 - b^2

= 100^2 - 4^2

= 10000 - 16

= 9984

Q4. Find (17 × 21) using an identity.

17 × 21 = (19 - 2)(19 + 2)

= 19^2 - 2^2

= 361 - 4

= 357

Q5. Factorise (25a^2 - 30ab + 9b^2).

25a^2 - 30ab + 9b^2

= (5a)^2 - 2(5a)(3b) + (3b)^2

= (5a - 3b)^2

Q6. Factorise (36s^2 - 49t^2).

Use:

a^2 - b^2 = (a + b)(a - b)

36s^2 - 49t^2 = (6s)^2 - (7t)^2

= (6s + 7t)(6s - 7t)

Rational Expressions Class 9 Important Questions

Rational expressions class 9 questions require factorisation first. Common factors can be cancelled only when the denominator is not zero.

Q1. Simplify ((x^2 - 7x + 12)/(5x^2 + 5x - 100)).

Factor the numerator:

x^2 - 7x + 12 = (x - 3)(x - 4)

Factor the denominator:

5x^2 + 5x - 100 = 5(x^2 + x - 20)

= 5(x + 5)(x - 4)

So:

(x^2 - 7x + 12)/(5x^2 + 5x - 100) ((x - 3)(x - 4))/(5(x + 5)(x - 4))

Cancel ((x - 4)).

= (x - 3)/(5(x + 5))

Q2. Simplify ((4x^2 + 4x + 1)/(4x^2 - 1)).

4x^2 + 4x + 1 = (2x + 1)^2

4x^2 - 1 = (2x + 1)(2x - 1)

So:

(4x^2 + 4x + 1)/(4x^2 - 1) ((2x + 1)^2)/((2x + 1)(2x - 1))

= (2x + 1)/(2x - 1)

Q3. Simplify ((s^3 + 125t^3)/(s^2 - 2st + 25t^2)).

Use:

a^3 + b^3 = (a + b)(a^2 - ab + b^2)

s^3 + 125t^3 = s^3 + (5t)^3

= (s + 5t)(s^2 - 5st + 25t^2)

This expression cannot cancel with (s^2 - 2st + 25t^2).

So, it remains:

((s + 5t)(s^2 - 5st + 25t^2))/(s^2 - 2st + 25t^2)

Application-Based Questions from Exploring Algebraic Identities Class 9

Application questions test whether students can turn a situation into an algebraic expression. These questions are useful for school exams and HOTS practice.

Q1. A rectangle has sides (x + 3) and (x + 4). Find its area.

Area of rectangle:

= (x + 3)(x + 4)

= x^2 + 4x + 3x + 12

= x^2 + 7x + 12

So, the area is (x^2 + 7x + 12).

Q2. The area of a rectangle is (x^2 + 8x + 15). Find possible length and breadth.

Factorise:

x^2 + 8x + 15

We need two numbers whose sum is 8 and product is 15.

The numbers are 3 and 5.

x^2 + 8x + 15 = (x + 3)(x + 5)

So, possible length and breadth are (x + 3) and (x + 5).

Q3. A rectangular pool has breadth 4 m less than its length and area 96 sq. m. Find its dimensions.

Let the length be (x) m.

Breadth = (x - 4) m.

Area:

x(x - 4) = 96

x^2 - 4x - 96 = 0

Factorise:

x^2 - 12x + 8x - 96 = 0

x(x - 12) + 8(x - 12) = 0

(x - 12)(x + 8) = 0

So, (x = 12) or (x = -8).

Length cannot be negative.

Length = 12 m

Breadth = 12 − 4 = 8 m

Q4. A square playground has side 40 m. A path of width (s) m is made around it. Find the area of the path.

Side of outer square:

40 + 2s

Area of outer square:

(40 + 2s)^2

Area of playground:

40^2 = 1600

Area of path:

(40 + 2s)^2 - 1600

Expand:

1600 + 160s + 4s^2 - 1600

= 160s + 4s^2

So, area of path is:

4s^2 + 160s

Long Answer Questions from Class 9 Maths Chapter 4

Long answers should show the identity, substitution and final result. For proof questions, write every expansion step clearly.

Q1. Explain how ((a + b)^2 = a^2 + 2ab + b^2) can be visualised geometrically.

Take a square of side (a + b).

Its total area is:

(a + b)^2

Now divide the square into four parts.

One square has side (a), so its area is (a^2).

Another square has side (b), so its area is (b^2).

Two rectangles have dimensions (a) and (b), so each has area (ab).

Total area:

a^2 + ab + ab + b^2

= a^2 + 2ab + b^2

Hence:

(a + b)^2 = a^2 + 2ab + b^2

Q2. Prove the pattern for three consecutive square numbers.

Let the three consecutive numbers be:

n - 1, n, n + 1

Their squares are:

(n - 1)^2, n^2, (n + 1)^2

Add the smallest and largest squares:

(n - 1)^2 + (n + 1)^2

= n^2 - 2n + 1 + n^2 + 2n + 1

= 2n^2 + 2

Now subtract twice the middle square:

2n^2 + 2 - 2n^2 = 2

So, the result is always 2.

Q3. Derive the identity ((a + b)^3).

(a + b)^3 = (a + b)(a + b)^2

Use:

(a + b)^2 = a^2 + 2ab + b^2

So:

(a + b)^3 = (a + b)(a^2 + 2ab + b^2)

= a(a^2 + 2ab + b^2) + b(a^2 + 2ab + b^2)

= a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3

= a^3 + 3a^2b + 3ab^2 + b^3

Q4. Use identities to prove (x^3 - y^3 = (x - y)(x^2 + xy + y^2)).

Expand the right side:

(x - y)(x^2 + xy + y^2)

= x(x^2 + xy + y^2) - y(x^2 + xy + y^2)

= x^3 + x^2y + xy^2 - x^2y - xy^2 - y^3

Cancel like terms:

= x^3 - y^3

Hence:

x^3 - y^3 = (x - y)(x^2 + xy + y^2)

Q5. Prove that if (a + b + c = 5) and (ab + bc + ca = 10), then (a^3 + b^3 + c^3 - 3abc = -25).

Use the identity:

a^3 + b^3 + c^3 - 3abc

= (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)

We know:

(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

Given:

a + b + c = 5

ab + bc + ca = 10

So:

25 = a^2 + b^2 + c^2 + 2(10)

25 = a^2 + b^2 + c^2 + 20

a^2 + b^2 + c^2 = 5

Now:

a^2 + b^2 + c^2 - ab - bc - ca = 5 - 10 = -5

Therefore:

a^3 + b^3 + c^3 - 3abc = 5(-5)

= -25

Useful Resources for Class 9 Maths