Important Questions Class 8 Maths Chapter 9 Algebraic Expressions and Identities

Important class 8 maths questions for chapter 9 Algebraic expressions and Identities will help students to get better prepared for CBSE class 8 exam and develop problem-solving skills. These algebraic expressions and identities questions not only cover NCERT questions but also other variations of questions to help class 8 students get acquainted with a wide range of questions.

Also Check:

• Important 2 Marks Questions for CBSE 8th Maths
• Important 3 Marks Questions for CBSE 8th Maths
• Important 4 Marks Questions for CBSE 8th Maths

Algebraic Expressions and Identities Important Questions For Class 8 (Chapter 9)

The algebraic identities and expressions questions given here include different short answer type and long answer type questions. It should be noted that the basic concepts of algebraic identities are required to solve most of the questions.

Short Answer Type Questions:

1. Using suitable algebraic identity, solve 1092²

Solution:

Use the algebraic identity: (a + b)² = a² + 2ab + b²

Now, 1092 = 1000 + 92

So, 1092² = (1000 + 92)²

(1000 + 92)² = ( 1000 )² + 2 × 1000 × 92 + ( 92 )²

= 1000000 + 184000 + 8464

Thus, 1092² = 1192464.

2. Identify the type of expressions:

(i) x²y + xy²

(ii) 564xy

(iii) -8x + 4y

(iv) x² + x + 7

(iv) xy + yz + zp + px + 9xy

Solution:

(i) x²y + xy² = Binomial

(ii) 564xy = Monomial

(iii) -8x + 4y = Binomial

(iv) x² + x + 7 = Trinomial

(iv) xy + yz + zp + px + 9xy = Polynomial

3. Identify terms and their coefficients from the following expressions:

(i) 6x²y² – 9x²y²z²+ 4z²

(ii) 3xyz – 8y

(iii) 6.1x – 5.9xy + 2.3y

Solution:

(i) 6x²y² – 9x²y²z²+ 4z²

Terms = 6x²y², -9x²y²z², and 4z²

Coefficients = 6, -9, and 4

(ii) 3xyz – 8y

Terms = 3xyz, and -8y

Coefficients = 3, and -8

(iii) 6.1x – 5.9xy + 2.3y

Terms = 6.1x, – 5.9xy, and 2.3y

Coefficients = 6.1, – 5.9 and 2.3

4. Find the area of a square with side 5x²y

Solution:

Given that the side of square = 5x²y

Area of square = side² = (5x²y)² = 25x⁴y²

5. Calculate the area of a rectangle whose length and breadths are given as 3x²y m and 5xy² m respectively.

Solution:

Given,

Length = 3x²y m

Breadth = 5xy² m

Area of rectangle = Length × Breadth

= (3x²y × 5xy²) = (3 × 5) × x²y × xy² = 15x³y³ m²

Long Answer Type Questions:

6. Simplify the following expressions:

(i) (x + y + z)(x + y – z)

(ii) x²(x – 3y²) – xy(y² – 2xy) – x(y³ – 5x²)

(iii) 2x²(x + 2) – 3x (x² – 3) – 5x(x + 5)

Solution:

Notes: “+” × “+” = “+”, “-” × “-” = “+”, and “+” × “-” = “-”.

(i) (x + y + z)(x + y – z)

= x² + xy – xz + yx + y² – yz + zx + zy – z²

Add similar terms like xy and yx, xz and zx, and yz and zy. Then simplify and rearrange.

= x² + y² – z² + 2xy

(ii) x²(x – 3y²) – xy(y² – 2xy) – x(y³ – 5x²)

= x³ – 3x²y² – xy³ + 2x²y² – xy³ + 5x³

Now, add the similar terms and rearrange.

= x³ + 5x³ – 3x²y² + 2x²y² – xy³ – xy³

= 6x³ – x²y² – 2xy³

(iii) 2x²(x + 2) – 3x (x² – 3) – 5x(x + 5)

= 2x³ + 4x² – 3x³ + 9x – 5x² – 25x

= 2x³ – 3x³ – 5x² + 4x² + 9x – 25x

= -x³ – x² – 16x

7. Add the following polynomials.

(i) x + y + xy, x – z + yx, and z + x + xz

(ii) 2x²y²– 3xy + 4, 5 + 7xy – 3x²y², and 4x²y² + 10xy

(iii) -3a²b², (–5/2) a²b², 4a²b², and (⅔) a²b²

Solution:

(i) x + y + xy, x – z + yx, and z + x + xz

= (x + y + xy) + (x – z + yx) + (z + x + xz)

= x + y + xy + x – z + yx + z + x + xz

Add similar elements and rearrange.

= 2xy + xz + 3x + y

(ii) 2x²y²– 3xy + 4, 5 + 7xy – 3x²y², and 4x²y² + 10xy

= (2x²y²– 3xy + 4) + (5 + 7xy – 3x²y²) + (4x²y² + 10xy)

= 2x²y² – 3xy + 4 + 5 + 7xy – 3x²y² + 4x²y² + 10xy

Add similar elements and rearrange.

= 3x²y² + 14xy + 9

(iii) -3a²b², (–5/2) a²b², 4a²b², and (⅔) a²b²

8. Subtract the following polynomials.

(i) (7x + 2) from (-6x + 8)

(ii) 3xy + 5yz – 7xz + 1 from -4xy + 2yz – 2xz + 5xyz + 1

(iii) 2x²y²– 3xy + 4 from 4x²y² + 10xy

Solution:

(i) (7x + 2) from (-6x + 8)

= (-6x + 8) – (7x + 2)

= -6x + 8 – (7x + 2)

= -6x + 8 – 7x – 2

= -13x + 6

(ii) 3xy + 5yz – 7xz + 1 from -4xy + 2yz – 2xz + 5xyz + 1

(iii) 2x²y²– 3xy + 4 from 4x²y² + 10xy

9. Calculate the volume of a cuboidal box whose dimensions are 5x × 3x² × 7x⁴

Solution:

Given,

Length = 5x

Breadth = 3x²

Height = 7x⁴

Volume of cuboid = Length × Breadth × Height

= 5x × 3x² × 7x⁴

Multiply 5, 3, and 7

= 105xx²x⁴

Use exponents rule: x^a × x^b = x^(a+b)

So, 105xx²x⁴ = 105x¹⁺²⁺⁴ = 105x⁷

10. Simplify 7x²(3x – 9) + 3 and find its values for x = 4 and x = 6

Solution:

7x²(3x – 9) + 3

Solve for 7x²(3x – 9)

= (7x² × 3x)  – (7x² × 9) (using distributive law: a(b – c) = ab – ac)

= 21x³ – 63x²

So, 7x²(3x – 9) + 3

= 21x³ – 63x² + 3

Now, for x = 4,

21x³ – 63x² + 3

= 21 × 4³ – 63 × 4² + 3

= 1344 – 1008 + 3

= 336 + 3 = 339

Now, for x = 6,

21x³ – 63x²

= 21 × 6³ – 63 × 6² + 3

= 2268 + 3

= 2271

Class 8 Maths Chapter 9 Extra Questions

1. What is the sum of ab, a+b  and b+ab?
(a) 2ab +2a +b
(b) 2ab+a+b
(c) 2ab+a+2b

2. Give the statement for the expression 2x-9
(a) 9 subtracted from x and multiplied by 2
(b) three of x minus 9
(c) 2x subtracted from 9

3. Give examples for each of:
(i) Monomials
(ii) Binomials
(iii) Trinomials

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