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

Important Questions Class 9 Maths Chapter 4 Exploring Algebraic Identities

This page gives Important Questions Class 9 Maths Chapter 4 Exploring Algebraic Identities. The chapter is about algebraic identities and factorisation. You will use identities like (a + b + c)², a² − b² and a³ + b³ + c³ − 3abc to factorise and simplify expressions.

Some questions use algebra tiles, area of a path, a pool and cuboid volumes, so you can see how identities help in real life. Every answer is solved step by step in easy words. Try each question yourself first, then check the steps.

Q1. A rectangular arrangement is formed using algebra tiles representing x², x and 1. The arrangement contains one x² tile, seven x tiles and twelve unit tiles. Using the concept of algebraic tiling, determine the length and breadth of the rectangle and hence factorise the expression x² + 7x + 12.

Answer: The given algebraic expression is:

x² + 7x + 12

To form a rectangle using algebra tiles, we need to arrange the seven x tiles and twelve unit tiles so that they form a complete rectangle along with the x² tile.

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

3 × 4 = 12

3 + 4 = 7

Splitting the middle term,

x² + 7x + 12 = x² + 3x + 4x + 12

Grouping the terms,

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

= (x + 3)(x + 4)

Thus, the dimensions of the rectangle are:

Length: x + 4

Breadth: x + 3

Hence, the factorised form is:

(x + 3)(x + 4)

Q2. The expression 9p² + 4q² + 25r² − 12pq + krq + 30pr is a perfect square of a trinomial. Find the value of k and hence factorise the expression.

Answer: Given expression:

9p² + 4q² + 25r² − 12pq + krq + 30pr

Since the expression is a perfect square, we compare it with the identity:

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

The square terms can be written as:

9p² = (3p)²

4q² = (−2q)²

25r² = (5r)²

Also,

−12pq = 2(3p)(−2q)

30pr = 2(3p)(5r)

Therefore, the trinomial must be:

3p − 2q + 5r

The remaining cross-product term is:

2(−2q)(5r) = −20qr

Comparing this with krq, we get:

k = −20

Substituting the value of k, the expression becomes:

9p² + 4q² + 25r² − 12pq − 20qr + 30pr

Using the identity,

= (3p − 2q + 5r)²

Hence,

k = −20

Factorised form:

(3p − 2q + 5r)²

Q3. Factorise the following expression using an appropriate algebraic identity: 4x² + 9y² + z² − 12xy + 6yz − 4xz

Answer: Given expression:

4x² + 9y² + z² − 12xy + 6yz − 4xz

Rearranging the terms,

= (2x)² + (−3y)² + (−z)² + 2(2x)(−3y) + 2(−3y)(−z) + 2(2x)(−z)

Using the identity,

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

We get,

= (2x − 3y − z)²

Hence, the factorised form is:

(2x − 3y − z)²

Q4. By factoring the expression, check that n³ − n is always divisible by 6 for all natural numbers n. Give reasons.

Answer: We have: n³ − n

= n(n² − 1)

= n(n − 1)(n + 1)

So, n³ − n = n(n − 1)(n + 1)

Thus, n³ − n is the product of three consecutive natural numbers: (n − 1), n, (n + 1)

Among any three consecutive natural numbers:

one is always divisible by 3

at least one is always even, so divisible by 2

Therefore, their product is always divisible by: 2 × 3 = 6

Hence, n³ − n is always divisible by 6 for all natural numbers n.

Q5. If a + b + c = 5 and ab + bc + ca = 10, then prove that a³ + b³ + c³ − 3abc = −25.

Answer: Given:

a + b + c = 5

ab + bc + ca = 10

First finding a² + b² + c² using:

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

So, 5² = a² + b² + c² + 2(10)

⇒ 25 = a² + b² + c² + 20

⇒ a² + b² + c² = 5

Now, a² + b² + c² − ab − bc − ca = 5 − 10 = −5

Using the identity: a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

a³ + b³ + c³ − 3abc

= (5)(−5)

= −25

Hence proved a³ + b³ + c³ − 3abc = −25.

Q6. If both x − 2 and x − 1/2 are factors of px² + 5x + r, show that p = r.

Answer: Since x − 2 and x − 1/2 are factors, the quadratic polynomial can be written as:

px² + 5x + r = k(x − 2)(x − 1/2)

Now, (x − 2)(x − 1/2) = x² − (1/2)x − 2x + 1 = x² − (5/2)x + 1

⇒ px² + 5x + r = k[x² − (5/2)x + 1]

Comparing the coefficients on both sides, we get:

Coefficient of x² gives: k = p

Constant term gives: k = r

Therefore, p = r

Hence proved.

Q7. A rectangular pool has area 2x² + 7x + 3 square hastas. If its width is 2x + 1 hastas, find its length. Hasta was a unit used to measure length.

Answer: Area of pool = 2x² + 7x + 3

Width = 2x + 1

Length = Area/Width

= (2x² + 7x + 3)/(2x + 1)

= (2x² + 6x + x + 3)/(2x + 1)

= [2x(x + 3) + 1(x + 3)]/(2x + 1)

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

= x + 3

Therefore, the length of the pool is x + 3 hastas.

Q8. If a number plus its reciprocal equals 10/3, find the number.

Answer: Let the number be x.

Then, according to question: x + 1/x = 10/3

Multiply both sides by 3x:

3x² + 3 = 10x

⇒ 3x² − 10x + 3 = 0

⇒ 3x² − 9x − x + 3 = 0

⇒ 3x(x − 3) − 1(x − 3) = 0

⇒ (3x − 1)(x − 3) = 0

So, 3x − 1 = 0 or x − 3 = 0

Hence, x = 1/3 or x = 3

Therefore, the number is 3 or 1/3.

Q9. The village playground is shaped as a square of side 40 metres. A path of width s metres is created around the playground for people to walk. Find an expression for the area of the path in terms of s.

Answer: Side of playground = 40 m

Since a path of width s metres is made all around the outside, the side of the outer square becomes:

40 + 2s

Area of outer square = (40 + 2s)²

Area of playground = 40² = 1600

Therefore, area of the path

= (40 + 2s)² − 1600

= (1600 + 160s + 4s²) − 1600

= 4s² + 160s

Hence, the required expression: Area of path = 4s² + 160s square metres.

Q10. Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units.

(i) 6a² − 24b²

(ii) 3ps² − 15ps + 12p

Answer: (i) 6a² − 24b²

Take common factor:

6a² − 24b²

= 6(a² − 4b²)

= 6[a² − (2b)²]

= 6(a + 2b)(a − 2b) [Using a² − b² = (a + b)(a − b)]

So possible dimensions are 6, (a + 2b) and (a − 2b).

(ii) 3ps² − 15ps + 12p

= 3p(s² − 5s + 4) [Taking common factor]

= 3p(s² − 4s − 1s + 4)

= 3p[s(s − 4) − 1(s − 4)]

= 3p(s − 1)(s − 4)

So, the possible dimensions are 3p, (s − 1) and (s − 4).

Q11. Try to simplify the following rational expression: (36s² − 12st + t²)/(t² + 2ts − 48s²) = (6s − t)²/(− + ... + −).

Answer: Given,

(36s² − 12st + t²)/(t² + 2ts − 48s²)

Factorise the numerator:

36s² − 12st + t² = (6s − t)²

Factorise the denominator:

t² + 2ts − 48s²

We need two terms whose product is −48s² and whose sum is 2s.

These terms are 8s and −6s.

So,

t² + 2ts − 48s² = (t + 8s)(t − 6s)

Since,

t − 6s = −(6s − t)

we get,

(6s − t)²/[(t + 8s)(t − 6s)]

= (6s − t)²/[−(t + 8s)(6s − t)]

Cancelling the common factor (6s − t),

= −(6s − t)/(t + 8s)

Q12. We already know that x² − y² = (x − y)(x + y). Further, we have verified that x³ − y³ = (x − y)(x² + xy + y²). Observe that x − y is a common factor of x² − y² and x³ − y³. Do you think x − y is also a factor of x⁴ − y⁴? Note that x⁴ − y⁴ = (x²)² − (y²)² = (x² − y²)(x² + y²). Can you see how x − y is a factor of x⁴ − y⁴? How about x⁵ − y⁵? Does this also have x − y as a factor?

Answer: Yes, x − y is a factor of both x⁴ − y⁴ and x⁵ − y⁵.

For x⁴ − y⁴,

x⁴ − y⁴ = (x²)² − (y²)²

Using the identity,

a² − b² = (a − b)(a + b)

we get,

x⁴ − y⁴ = (x² − y²)(x² + y²)

Also,

x² − y² = (x − y)(x + y)

Therefore,

x⁴ − y⁴ = (x − y)(x + y)(x² + y²)

So, x − y is a factor of x⁴ − y⁴.

For x⁵ − y⁵,

x⁵ − y⁵ = (x − y)(x⁴ + x³y + x²y² + xy³ + y⁴)

Therefore, x − y is also a factor of x⁵ − y⁵.

Related Study Material on Extramarks

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• Important Questions for Class 9 Maths Chapter 2 Introduction to Linear Polynomials
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