Algebraic Expressions and Identities Class 8 Extra Questions Maths Chapter 9

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

Extra Questions for Class 8 Maths

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