Class 9 Maths Chapter 4 Important Questions with Solutions – Exploring Algebraic Identities (Ganita Manjari)

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Last Updated on November 7, 2026 by XAM CONTENT

Chapter 4 Practice

Practice these Class 9 Maths Chapter 4 Important Questions from Exploring Algebraic Identities in the new NCERT Ganita Manjari. The set covers algebraic identities, visual interpretation, expansion, factorisation, cubes and rational expressions. Attempt each question first, then open the solution to check your method.

Class
CBSE Class 9
Subject
Mathematics
Chapter
4 — Exploring Algebraic Identities
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Key Concept

An algebraic identity is true for all permissible values of its variables. Identities can be used in both directions: to expand algebraic expressions and to recognise patterns for factorisation.

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Quick Revision

Important Algebraic Identities

Revise these identities before attempting the questions.
Identity Formula
Square of a sum \((a+b)^2=a^2+2ab+b^2\)
Square of a difference \((a-b)^2=a^2-2ab+b^2\)
Difference of two squares \((a+b)(a-b)=a^2-b^2\)
Square of three terms \((a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\)
Product with common first term \((x+a)(x+b)=x^2+(a+b)x+ab\)
Cube of a sum \((a+b)^3=a^3+3a^2b+3ab^2+b^3\)
Cube of a difference \((a-b)^3=a^3-3a^2b+3ab^2-b^3\)
Sum of cubes \(a^3+b^3=(a+b)(a^2-ab+b^2)\)
Difference of cubes \(a^3-b^3=(a-b)(a^2+ab+b^2)\)
⚠️
Common Mistake

Remember that \((a+b)^2\neq a^2+b^2\). The middle term \(2ab\) is essential. Similarly, when factorising, check both the sum and the product of the numbers used to split the middle term.

✍️
Practice Set

Class 9 Maths Chapter 4 Important Questions with Solutions

Conceptual, numerical, factorisation and higher-order questions from the complete chapter.
12
Questions
🧩
Important Questions
Chapter 4 — Exploring Algebraic Identities
📚 Class 9 Maths 🧩 Algebraic Identities 🧠 Concept + Application
1

Which statement is an algebraic identity?

Easy
A\(x^2-1=24\)
B\((x+3)^2=x^2+6x+9\)
C\(2x+5=17\)
D\(x^2=49\)
Explanation

An identity is true for every permissible value of its variable. Expanding the left side gives

\[ (x+3)^2=x^2+2(x)(3)+3^2=x^2+6x+9. \]

Answer: B.

2

Take three consecutive natural numbers \((n-1),n,(n+1)\). Prove that the sum of the squares of the first and third numbers minus twice the square of the middle number is always 2.

Medium
Solution

The required expression is

\[ (n-1)^2+(n+1)^2-2n^2. \] \[ =n^2-2n+1+n^2+2n+1-2n^2 \] \[ =2. \]

The variable \(n\) disappears completely, so the value is 2 for every natural number \(n\).

3

A square has side \((a+b)\). Explain how its area gives the identity \((a+b)^2=a^2+2ab+b^2\). Hence expand \((3x+2y)^2\).

Medium
Solution

A square of side \((a+b)\) can be divided into one square of area \(a^2\), one square of area \(b^2\), and two rectangles of area \(ab\) each.

\[ (a+b)^2=a^2+ab+ab+b^2=a^2+2ab+b^2. \]
Geometric representation of the identity a plus b whole square

Now take \(a=3x\) and \(b=2y\):

\[ (3x+2y)^2=(3x)^2+2(3x)(2y)+(2y)^2 \] \[ \boxed{(3x+2y)^2=9x^2+12xy+4y^2}. \]
4

Calculate \(103^2\) mentally using a suitable algebraic identity.

Easy
Solution

Write \(103=100+3\) and use \((a+b)^2=a^2+2ab+b^2\):

\[ 103^2=(100+3)^2 \] \[ =10000+2(100)(3)+9 \] \[ =10000+600+9 \] \[ \boxed{103^2=10609}. \]
5

Factorise \(25p^2+30pq+9q^2\) using an algebraic identity.

Easy
Solution

Observe that

\[ 25p^2=(5p)^2,\qquad 9q^2=(3q)^2 \]

and the middle term is

\[ 2(5p)(3q)=30pq. \]

Therefore,

\[ \boxed{25p^2+30pq+9q^2=(5p+3q)^2}. \]
6

The value of \((5x+7y)(5x-7y)\) is:

Easy
A\(25x^2-49y^2\)
B\(25x^2+49y^2\)
C\(25x^2-70xy+49y^2\)
D\(25x^2+70xy+49y^2\)
Explanation

Use \((a+b)(a-b)=a^2-b^2\), with \(a=5x\) and \(b=7y\):

\[ (5x+7y)(5x-7y)=(5x)^2-(7y)^2=25x^2-49y^2. \]

Answer: A.

7

Expand and simplify \((x+2y-3z)^2\).

Medium
Solution

Use

\[ (a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca. \]

Here \(a=x\), \(b=2y\), \(c=-3z\). Therefore,

\[ (x+2y-3z)^2 =x^2+4y^2+9z^2+4xy-12yz-6xz. \]

Hence,

\[ \boxed{x^2+4y^2+9z^2+4xy-6xz-12yz}. \]
8

Factorise \(x^2+9x+20\). Explain how the factors can be interpreted as the side lengths of a rectangle formed using algebra tiles.

Medium
Solution

We need two numbers whose sum is 9 and product is 20. The numbers are 4 and 5.

\[ x^2+9x+20=x^2+4x+5x+20 \] \[ =x(x+4)+5(x+4) \] \[ \boxed{x^2+9x+20=(x+4)(x+5)}. \]

Thus, an algebra-tile rectangle representing the expression has side lengths \((x+4)\) and \((x+5)\).

Algebra tile rectangle representing x squared plus 9x plus 20
9

Factorise \(6x^2+7x+2\) by splitting the middle term.

Medium
Solution

The product of the coefficient of \(x^2\) and the constant term is

\[ 6\times2=12. \]

We need two numbers whose product is 12 and sum is 7. They are 3 and 4.

\[ 6x^2+7x+2=6x^2+3x+4x+2 \] \[ =3x(2x+1)+2(2x+1) \] \[ \boxed{6x^2+7x+2=(3x+2)(2x+1)}. \]
10

(i) Expand \((2p-3q)^3\).
(ii) Factorise \(x^3-27\).

Challenging
Solution

(i) Use

\[ (a-b)^3=a^3-3a^2b+3ab^2-b^3. \]

Taking \(a=2p\) and \(b=3q\),

\[ (2p-3q)^3 =8p^3-36p^2q+54pq^2-27q^3. \]

(ii) Use \(a^3-b^3=(a-b)(a^2+ab+b^2)\):

\[ x^3-27=x^3-3^3 \] \[ \boxed{x^3-27=(x-3)(x^2+3x+9)}. \]
11

Three numbers \(x,y,z\) satisfy \(x+y+z=10\), \(xyz=25\), and \(x^2+y^2+z^2=38\). Find \(x^3+y^3+z^3\).

Challenging
Solution

First use

\[ (x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx). \]

Therefore,

\[ 100=38+2(xy+yz+zx) \] \[ xy+yz+zx=31. \]

Now use

\[ x^3+y^3+z^3-3xyz =(x+y+z)(x^2+y^2+z^2-xy-yz-zx). \]

Substituting the given values,

\[ x^3+y^3+z^3-75=10(38-31)=70. \] \[ \boxed{x^3+y^3+z^3=145}. \]
12

Simplify the rational expression \(\frac{x^2-5x+6}{2x^2+2x-12}\) and state the values of \(x\) for which the original expression is undefined.

Challenging
Solution

Factor the numerator:

\[ x^2-5x+6=(x-2)(x-3). \]

Factor the denominator:

\[ 2x^2+2x-12=2(x^2+x-6) \] \[ =2(x+3)(x-2). \]

Therefore,

\[ \frac{(x-2)(x-3)}{2(x+3)(x-2)} =\boxed{\frac{x-3}{2(x+3)}}. \]

However, the restrictions must come from the original denominator:

\[ 2(x+3)(x-2)\neq0. \]

Hence, \(\boxed{x\neq2,-3}\).

✅
Exam Tip

Before expanding a long expression, look for a standard identity. Before factorising, first take out any common factor and then check whether the remaining expression matches a perfect-square, difference-of-squares, cube, or middle-term-splitting pattern.

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FAQs

Class 9 Maths Chapter 4 Important Questions — FAQs

What is an algebraic identity?

An algebraic identity is an equality that is true for all permissible values of its variables, unlike an equation that may be true only for particular values.

What is the identity for \((a+b)^2\)?

The identity is \((a+b)^2=a^2+2ab+b^2\). The term \(2ab\) comes from two equal rectangular parts in its geometric interpretation.

How are algebraic identities useful in factorisation?

Expansion identities can be read in reverse. For example, \(a^2+2ab+b^2\) can immediately be factorised as \((a+b)^2\).

How do we factorise \(x^2+bx+c\)?

Find two numbers whose sum is \(b\) and whose product is \(c\). These numbers determine the two binomial factors.

What should be remembered when simplifying rational algebraic expressions?

Factor the numerator and denominator completely, cancel only common factors, and always retain the restrictions obtained from the original denominator.

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