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 1092 ^{2}**

**Solution:**

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

Now, 1092 = 1000 + 92

So, 1092^{2} = (1000 + 92)^{2}

(1000 + 92)^{2} = ( 1000 )² + 2 × 1000 × 92 + ( 92 )²

= 1000000 + 184000 + 8464

Thus, 1092^{2} = 1192464.

**2. Identify the type of expressions:**

**(i) x ^{2}y + xy^{2} **

**(ii) 564xy **

**(iii) -8x + 4y **

**(iv) x ^{2} + x + 7 **

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

**Solution:**

(i) x^{2}y + xy^{2} = Binomial

(ii) 564xy = Monomial

(iii) -8x + 4y = Binomial

(iv) x^{2} + x + 7 = Trinomial

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

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

**(i) 6x ^{2}y^{2} – 9x^{2}y^{2}z^{2}+ 4z^{2}**

**(ii) 3xyz – 8y**

**(iii) 6.1x – 5.9xy + 2.3y**

**Solution:**

(i) 6x^{2}y^{2} – 9x^{2}y^{2}z^{2}+ 4z^{2}

Terms = 6x^{2}y^{2}, -9x^{2}y^{2}z^{2}, and 4z^{2}

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 5x ^{2}y**

**Solution:**

Given that the side of square = 5x^{2}y

Area of square = side^{2} = (5x^{2}y)^{2} = 25x^{4}y^{2}

**5. Calculate the area of a rectangle whose length and breadths are given as 3x ^{2}y m and 5xy^{2} m respectively.**

**Solution:**

Given,

Length = 3x^{2}y m

Breadth = 5xy^{2} m

Area of rectangle = Length × Breadth

= (3x^{2}y × 5xy^{2}) = (3 × 5) × x^{2}y × xy^{2} = 15x^{3}y^{3} m^{2}

### Long Answer Type Questions:

**6. Simplify the following expressions: **

**(i) (x + y + z)(x + y – z)**

**(ii) x ^{2}(x – 3y^{2}) – xy(y^{2} – 2xy) – x(y^{3} – 5x^{2})**

**(iii) 2x ^{2}(x + 2) – 3x (x^{2} – 3) – 5x(x + 5)**

**Solution:**

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

(i) (x + y + z)(x + y – z)

= x^{2} + xy – xz + yx + y^{2} – yz + zx + zy – z^{2}

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

= x^{2} + y^{2} – z^{2} + 2xy

(ii) x^{2}(x – 3y^{2}) – xy(y^{2} – 2xy) – x(y^{3} – 5x^{2})

= x^{3} – 3x^{2}y^{2} – xy^{3} + 2x^{2}y^{2} – xy^{3} + 5x^{3}

Now, add the similar terms and rearrange.

= x^{3} + 5x^{3} – 3x^{2}y^{2} + 2x^{2}y^{2} – xy^{3} – xy^{3}

= 6x^{3} – x^{2}y^{2} – 2xy^{3}

(iii) 2x^{2}(x + 2) – 3x (x^{2} – 3) – 5x(x + 5)

= 2x^{3} + 4x^{2} – 3x^{3} + 9x – 5x^{2} – 25x

= 2x^{3} – 3x^{3} – 5x^{2} + 4x^{2} + 9x – 25x

= -x^{3} – x^{2} – 16x

**7. Add the following polynomials.**

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

**(ii) 2x ^{2}y^{2}– 3xy + 4, 5 + 7xy – 3x^{2}y^{2}, and 4x^{2}y^{2} + 10xy**

**(iii) -3a ^{2}b^{2}, (–5/2) a^{2}b^{2}, 4a^{2}b^{2}, and (⅔) a^{2}b^{2}**

**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) 2x^{2}y^{2}– 3xy + 4, 5 + 7xy – 3x^{2}y^{2}, and 4x^{2}y^{2} + 10xy

= (2x^{2}y^{2}– 3xy + 4) + (5 + 7xy – 3x^{2}y^{2}) + (4x^{2}y^{2} + 10xy)

= 2x^{2}y^{2 }– 3xy + 4 + 5 + 7xy – 3x^{2}y^{2 }+ 4x^{2}y^{2 }+ 10xy

Add similar elements and rearrange.

= 3x^{2}y^{2 }+ 14xy + 9

(iii) -3a^{2}b^{2}, (–5/2) a^{2}b^{2}, 4a^{2}b^{2}, and (⅔) a^{2}b^{2}

**8. Subtract the following polynomials.**

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

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

**(iii) 2x ^{2}y^{2}– 3xy + 4 from 4x^{2}y^{2} + 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

\(=-4xy+2yz-2xz+5xyz+1-\left(3xy+5yz-7xz+1\right)\\ =-4xy+2yz-2xz+5xyz+1-3xy-5yz+7xz-1\\ =5xz+5xyz-7xy-3yz\\\)(iii) 2x^{2}y^{2}– 3xy + 4 from 4x^{2}y^{2} + 10xy

**9. Calculate the volume of a cuboidal box whose dimensions are 5x × 3x ^{2} × 7x^{4}**

**Solution:**

Given,

Length = 5x

Breadth = 3x^{2}

Height = 7x^{4}

Volume of cuboid = Length × Breadth × Height

= 5x × 3x^{2} × 7x^{4}

Multiply 5, 3, and 7

= 105xx^{2}x^{4}

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

So, 105xx^{2}x^{4} = 105x^{1+2+4} = 105x^{7}

**10. Simplify 7x ^{2}(3x – 9) + 3 and find its values for x = 4 and x = 6**

**Solution:**

7x^{2}(3x – 9) + 3

Solve for 7x^{2}(3x – 9)

= (7x^{2} × 3x) – (7x^{2} × 9) (using distributive law: a(b – c) = ab – ac)

= 21x^{3} – 63x^{2}

So, 7x^{2}(3x – 9) + 3

= 21x^{3} – 63x^{2} + 3

Now, for x = 4,

21x^{3} – 63x^{2} + 3

= 21 × 4^{3} – 63 × 4^{2} + 3

= 1344 – 1008 + 3

= 336 + 3 = 339

Now, for x = 6,

21x^{3} – 63x^{2}

= 21 × 6^{3} – 63 × 6^{2} + 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

More Topics Related to Class 8 Algebraic Expressions And Identities:

Algebra Formulas | NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities |

Algebraic Expressions and Identities Class 8 Notes: Chapter 9 | Algebraic Identities For Class 8 |

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for the first question there is an easier method to solve-

use identity (a-b)^2=a^2-2*a*b+b^2

(1092)^2=(1100-8)^2

=(1100)^2-(2*1100*8)+(8)^2

=1210000-17600+64

=1192464

i used the same method

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