# 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.

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## 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.

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

Area of rectangle = Length × Breadth

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

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

(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

= 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

= 3x2y2 + 14xy + 9

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

$=-3a^2b^2+\left(-\frac{5}{2}\right)a^2b^2+4a^2b^2+\left(\frac{2}{3}\right)a^2b^2\\ =-3a^2b^2-\frac{5}{2}a^2b^2+4a^2b^2+\frac{2}{3}a^2b^2\\ =a^2b^2\left(-3-\frac{5}{2}+4+\frac{2}{3}\right)\\ =-\frac{5}{6}a^2b^2\\ =-\frac{5a^2b^2}{6}\\$

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

$=-4xy+2yz-2xz+5xyz+1-\left(3xy+5yz-7xz+1\right)\\ =-4xy+2yz-2xz+5xyz+1-3xy-5yz+7xz-1\\ =5xz+5xyz-7xy-3yz\\$

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

$=4x^2y^2+10xy-\left(2x^2y^2-3xy+4\right)\\ =4x^2y^2+10xy-2x^2y^2+3xy-4\\ =2x^2y^2+13xy-4\\$

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

Solution:

Given,

Length = 5x

Height = 7x4

Area of cuboid = Length × Breadth × Height

= 5x × 3x2 × 7x4

Multiply 5, 3, and 7

= 105xx2x4

Use exponents rule: xa × xb = 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

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1. 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

1. trinetra

i used the same method

2. very good efforts

3. Nikita Singh

Very nice 👌

4. Akshit Sayana

These are very good question