# Algebraic Expressions and Identities Class 8 Extra Questions Maths Chapter 9

Extra Questions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities

### Algebraic Expressions and Identities Class 8 Extra Questions Very Short Answer Type

Question 1.
Write two examples of each of
(i) Monomials
(ii) Binomials
(iii) Trinomials
Solution:
(i) Monomials:
(a) 3x
(b) 5xy2
(ii) Binomials:
(a) p + q
(b) -5a + 2b
(iii) Trinomials:
(a) a + b + c
(b) x2 + x + 2

Question 2.
Identify the like expressions.
5x, -14x, 3x2 + 1, x2, -9x2, xy, -3xy
Solution:
Like terms: 5x and -14x, x2 and -9x2, xy and -3xy

Question 3.
Identify the terms and their coefficients for each of the following expressions:
(i) 3x2y – 5x
(ii) xyz – 2y
(iii) -x – x2
Solution: Question 4.
Add: -3a2b2, –$$\frac { 5 }{ 2 }$$ a2b2, 4a2b2, $$\frac { 2 }{ 3 }$$ a2b2
Solution:  Question 5.
Add: 8x2 + 7xy – 6y2, 4x2 – 3xy + 2y2 and -4x2 + xy – y2
Solution: Question 6.
Subtract: (4x + 5) from (-3x + 7)
Solution:
(-3x + 7) – (4x + 5) = -3x + 7 – 4x – 5 = -3x – 4x + 7 – 5 = -7x + 2

Question 7.
Subtract: 3x2 – 5x + 7 from 5x2 – 7x + 9
Solution:
(5x2 – 7x + 9) – (3x2 – 5x + 7)
= 5x2 – 7x + 9 – 3x2 + 5x – 7
= 5x2 – 3x2 + 5x – 7x + 9 – 7
= 2x2 – 2x + 2

Question 8.
Multiply the following expressions:
(a) 3xy2 × (-5x2y)
(b) $$\frac { 1 }{ 2 }$$ x2yz × $$\frac { 2 }{ 3 }$$ xy2z × $$\frac { 1 }{ 5 }$$ x2yz
Solution: Question 9.
Find the area of the rectangle whose length and breadths are 3x2y m and 5xy2 m respectively.
Solution:
Length = 3x2y m, breadth = 5xy2 m
Area of rectangle = Length × Breadth = (3x2y × 5xy2) sq m = (3 × 5) × x2y × xy2 sq m = 15x3y3 sq m

Question 10.
Multiply x2 + 7x – 8 by -2y.
Solution: ### Algebraic Expressions and Identities Class 8 Extra Questions Short Answer Type

Question 11.
Simplify the following:
(i) a2 (b2 – c2) + b2 (c2 – a2) + c2 (a2 – b2)
(ii) x2(x – 3y2) – xy(y2 – 2xy) – x(y3 – 5x2)
Solution:
(i) a2 (b2 – c2) + b2 (c2 – a2) + c2 (a2 – b2)
= a2b2 – a2c2) + b2c2 – b2a2) + c2a2 – c2b2)
= 0
(ii) x2(x – 3y2) – xy(y2 – 2xy) – x(y3 – 5x2)
= x3 – 3x2y2 – xy3 + 2x2y2 – xy3 + 5x3
= x3 + 5x3 – 3x2y2 + 2x2y2 – xy3 – xy3
= 6x3 – x2y2 – 2xy3

Question 12.
Multiply (3x2 + 5y2) by (5x2 – 3y2)
Solution:
(3x2 + 5y2) × (5x2 – 3y2)
= 3x2(5x2 – 3y2) + 5y2(5x2 – 3y2)
= 15x4 – 9x2y2 + 25x2y2 – 15y4
= 15x4 + 16x2y2 – 15y4

Question 13.
Multiply (6x2 – 5x + 3) by (3x2 + 7x – 3)
Solution:
(6x2 – 5x + 3) × (3x2 + 7x – 3)
= 6x2(3x2 + 7x – 3) – 5x(3x2 + 7x – 3) + 3(3x2 + 7x – 3)
= 18x4 + 42x3 – 18x2 – 15x3 – 35x2 + 15x + 9x2 + 21x – 9
= 18x4 + 42x3 – 15x3 – 18x2 – 35x2 + 9x2 + 15x + 21x – 9
= 18x4 + 27x3 – 44x2 + 36x – 9

Question 14.
Simplify:
2x2(x + 2) – 3x (x2 – 3) – 5x(x + 5)
Solution:
2x2(x + 2) – 3x(x2 – 3) – 5x(x + 5)
= 2x3 + 4x2 – 3x3 + 9x – 5x2 – 25x
= 2x3 – 3x3 – 5x2 + 4x2 + 9x – 25x
= -x3 – x2 – 16x

Question 15.
Multiply x2 + 2y by x3 – 2xy + y3 and find the value of the product for x = 1 and y = -1.
Solution:
(x2 + 2y) × (x3 – 2xy + y3)
= x2(x3 – 2xy + y3) + 2y(x3 – 2xy + y3)
= x5 – 2x3y + x2y3 + 2x3y – 4xy2 + 2y4
= x5 + x2y3 – 4xy2 + 2y4
Put x = 1 and y = -1
= (1)5 + (1)2 (-1)3 – 4(1)(-1)2 + 2(-1)4
= 1 + (1) (-1) – 4(1)(1) + 2(1)
= 1 – 1 – 4 + 2
= -2

Question 16.
Using suitable identity find:
(i) 482 (NCERT Exemplar)
(ii) 962
(iii) 2312 – 1312
(iv) 97 × 103
(v) 1812 – 192 = 162 × 200 (NCERT Exemplar)
Solution: Question 17. Solution:  Question 18.
Verify that (11pq + 4q)2 – (11pq – 4q)2 = 176pq2 (NCERT Exemplar)
Solution:
LHS = (11pq + 4q)2 – (11pq – 4q)2 = (11pq + 4q + 11pq – 4q) × (11pq + 4q – 11pq + 4q)
[using a2 -b2 = (a – b) (a + b), here a = 11pq + 4q and b = 11 pq – 4q]
= (22pq) (8q)
= 176 pq2
= RHS.
Hence Verified.

Question 19.
Find the value of $$\frac { { 38 }^{ 2 }-{ 22 }^{ 2 } }{ 16 }$$, using a suitable identity. (NCERT Exemplar)
Solution: Question 20.
Find the value of x, if 10000x = (9982)2 – (18)2 (NCERT Exemplar)
Solution:
RHS = (9982)2 – (18)2 = (9982 + 18)(9982 – 18)
[Since a2 -b2 = (a + b) (a – b)]
= (10000) × (9964)
LHS = (10000) × x
Comparing L.H.S. and RHS, we get
10000x = 10000 × 9964
x = 9964                        +