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

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

. 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– 3xyz + 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

Answer= (⅚)a2b2 (Self Assessment)

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

= 5xz + 5xyz – 7xy – 3yz (Self Assessment)

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

= 21xy – 4 (Self Assessment)

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

Solution:

Given,

Length = 5x

Breadth = 3x2

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)

= 7 × 3x2x – 7 × 9x2

= 21x3 – 63x2

So, 7x2(3x – 9) + 3

= 21x3 – 63x2 + 3

Now, for x = 4,

21x3 – 63x2

= 21 × 43 – 63 × 42

= 1344 – 1008

= 336

Now, for x = 6,

21x3 – 63x2

= 21 × 63 – 63 × 62

= 2268