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Class 9 · Maths · Chapter 4 · NCERT Class 9 Maths

Exploring Algebraic Identities Class 9 Notes

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Chapter notes

CLASS 9 MATHS · CHAPTER 4 · STUDY NOTES

Exploring Algebraic Identities

Understand exploring algebraic identities through short explanations, labelled models and original practice. Read one topic, explain it in your own words, then check.

01

UNDERSTAND · APPLY · CHECK

An identity works for every allowed value

Tell an identity from an equation to solve.

The two middle rectangles matter

The two middle rectangles matterConcept model. a². ab. ab. b². Side a + b. For non-negative lengths a and b, the four areas total (a + b)². Distributivity extends the identity to all real a and b.a²ababb²Side a + b
Blue
a × a
Orange
Two separate a × b rectangles
Violet
b × b
For non-negative lengths a and b, the four areas total (a + b)². Distributivity extends the identity to all real a and b.
Identity
An equality true for every allowed value of its variables.

(x + 1)² = x² + 2x + 1 is an identity. x + 1 = 5 is an equation whose solution is x = 4.

Testing a few values can catch a false rule, but cannot prove a rule for all values. Distributivity explains why an identity must hold.

Take a moment. What do you think?

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Question

Is (a + b)² = a² + b² an identity?

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NCERT reference: chapter PDF pages 1, 3, 4.

02

UNDERSTAND · APPLY · CHECK

Count the two ab pieces

Explain and use (a + b)² = a² + 2ab + b².

The two middle rectangles matter

The two middle rectangles matterConcept model. a². ab. ab. b². Side a + b. For non-negative lengths a and b, the four areas total (a + b)². Distributivity extends the identity to all real a and b.a²ababb²Side a + b
Blue
a × a
Orange
Two separate a × b rectangles
Violet
b × b
For non-negative lengths a and b, the four areas total (a + b)². Distributivity extends the identity to all real a and b.
Binomial
An expression with two unlike terms.

Multiply (a + b)(a + b): the four products are a², ab, ba and b². The middle products combine to 2ab.

For (2x + 3)², treat 2x as a and 3 as b. The answer is 4x² + 12x + 9; the middle term uses both whole terms.

WORKED IDEA

A nearby square

52² = (50 + 2)² = 2500 + 200 + 4 = 2704

Choose an easy base number and keep the middle term.

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Question

Expand (x + 4)². Why does 8x appear?

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NCERT reference: chapter PDF pages 2, 3, 4, 5.

03

UNDERSTAND · APPLY · CHECK

A minus changes the middle term

Explain why the final b² remains positive.

Subtract two strips; add the overlap back

Subtract two strips; add the overlap backConcept model. (a − b)². ab. ab. b². Outer side a. For a ≥ b ≥ 0, removing two strips counts their b² overlap twice, so add that overlap back once. The algebraic identity itself holds for all real a and b.(a − b)²ababb²Outer side a
For a ≥ b ≥ 0, removing two strips counts their b² overlap twice, so add that overlap back once. The algebraic identity itself holds for all real a and b.
Square of a difference
(a − b)² = a² − 2ab + b².

Replace b by −b in the square-of-a-sum identity. The two middle products are negative, but (−b)² is positive.

For (3x − 2)², get 9x² − 12x + 4. A negative final term would be a sign error.

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Question

Calculate 48² using (50 − 2)².

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NCERT reference: chapter PDF pages 6, 7, 8.

04

UNDERSTAND · APPLY · CHECK

Every pair contributes twice

Expand the square of three terms systematically.

Nine regions explain the six cross products

Nine regions explain the six cross productsConcept model. a². ab. ac. ab. b². bc. ac. bc. c². Schematic, not to scale: the nine regions label the products, not equal a, b and c lengths. Every off-diagonal product occurs twice. The area interpretation uses positive lengths; algebra handles negative terms.a²abacabb²bcacbcc²
Schematic, not to scale: the nine regions label the products, not equal a, b and c lengths. Every off-diagonal product occurs twice. The area interpretation uses positive lengths; algebra handles negative terms.
Trinomial square
(a+b+c)² = a²+b²+c²+2ab+2ac+2bc.

Square each term, then add twice each pairwise product. There are three different pairs: ab, ac and bc.

For (x + 2y − 1)², the squares give x² + 4y² + 1; the pair terms give +4xy − 2x − 4y. Keep the sign of each complete term.

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Question

Which cross term comes from x and −1?

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NCERT reference: chapter PDF pages 8, 9, 10.

05

UNDERSTAND · APPLY · CHECK

A sum times a difference cancels the cross terms

Use a² − b² = (a + b)(a − b).

Remove a small square from a large one

Remove a small square from a large oneConcept model. a² − b². b². Outer square: side a. For a ≥ b ≥ 0, the remaining L-shaped area is a² − b² and can be rearranged to a rectangle with sides a+b and a−b.a² − b²b²Outer square: side a
For a ≥ b ≥ 0, the remaining L-shaped area is a² − b² and can be rearranged to a rectangle with sides a+b and a−b.
Difference of squares
The subtraction of two squared terms.

Multiply (a + b)(a − b): a² − ab + ab − b². The middle terms cancel, leaving a² − b².

This is different from (a − b)². For 9x² − 16, the square roots of the terms are 3x and 4, so the factors are (3x − 4)(3x + 4).

WORKED IDEA

Products around an easy centre

63 × 57 = (60 + 3)(60 − 3) = 3600 − 9 = 3591

The same offset above and below 60 lets us use the identity.

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Question

Why does a² + b² not factor as (a + b)(a − b)?

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NCERT reference: chapter PDF pages 10, 11.

06

UNDERSTAND · APPLY · CHECK

Find two numbers with the right sum and product

Read a quadratic area backwards to find its factors.

A rectangle explains the factors

A rectangle explains the factorsConcept model. x². 5x. 2x. 10. Width x + 5. Height x + 2. Non-negative x is used for the area model. The total x² + 7x + 10 factors as (x + 2)(x + 5).x²5x2x10Width x + 5Height x + 2
Non-negative x is used for the area model. The total x² + 7x + 10 factors as (x + 2)(x + 5).
Factorisation
Writing an expression as a product of simpler expressions.

x² + 7x + 10 needs two numbers whose sum is 7 and product is 10. They are 2 and 5, so it equals (x + 2)(x + 5).

For x² − x − 12, the pair is −4 and +3. Check: −4 + 3 = −1 and (−4) × 3 = −12. The factors are (x − 4)(x + 3).

  1. 1

    Look for a common factor first.

  2. 2

    Find a pair with the required sum and product.

  3. 3

    Split the middle term and group.

  4. 4

    Multiply your proposed factors to check.

Let’s look a little closer.

Go deeper: When the leading coefficient is not 1

For 2x² + 7x + 3, use a pair with product 2 × 3 = 6 and sum 7: 6 and 1. Split: 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).

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Question

Why would splitting 7x as x + 6x fail for x² + 7x + 10?

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NCERT reference: chapter PDF pages 5, 6, 11, 12, 13, 14, 15.

CHAPTER RECAP

An identity works for every allowed value → Count the two ab pieces → A minus changes the middle term → Every pair contributes twice → A sum times a difference cancels the cross terms → Find two numbers with the right sum and product

Learnijoy explanations and visuals based on NCERT Class IX Mathematics · Chapter 4 ↗. Exploring visuals and revealing answers do not record completion.

The rest of this chapter

Keep reading Exploring Algebraic Identities, free

  1. Locked: 1. A bigger cube contains more than two cubes
  2. Locked: 2. Factor cubes with a three-term partner
  3. Locked: 3. Join known identities to get a new result
  4. Locked: 4. Cancel factors, not pieces of a sum
  5. Locked: 5. Choose the algebra, then check the situation

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