RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

Class 8: RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

The document RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
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Question 1: Find each of the following product:
5x2 × 4x3

Answer 1: To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices. However, use of these laws are subject to their applicability in the given expressions.
In the present problem, to perform the multiplication, we can proceed as follows:
5x2×4x3=(5×4)×(x2×x3)5x2×4x3=5×4×x2×x3
=20x5=20x5                           ( am×an=am+nam×an=am+n)
Thus, the answer is 20x520x5. 

Question 2: Find each of the following product:
−3a2 × 4b4

Answer 2: To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, am×an=am+nam×an=am+n, wherever applicable.
We have:
3a2×4b4=(3×4)×(a2×b4)=12a2b4-3a2×4b4=-3×4×a2×b4=-12a2Thus, the answer is 12a2b4-12a2b4. 

Question 3: Find each of the following product:
(−5xy) × (−3x2yz)

Answer 3: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, am×an=am+nam×an=am+n, wherever applicable.
We have:
(5xy)×(3x2yz)={(5)×(3)}× (x×x2)×(y×y)×z=15× (x1+2)×(y1+1)×z=15x3y2z-5xy×-3x2yz=-5×-3× x×x2×y×y×z=15× x1+2×y1+1×z=15x3y2Thus, the answer is 15x3y2z15x3y2z. 

Question 4: Find each of the following product:

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

Answer 4: To multiply algebraic expressions, we use commutative and associative laws along with the the law of indices, that is, am×an=am+nam×an=am+n. 

We have: 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

Thus, the answer is 1/6 x3y2z216x3y2z2. 

Question 5: Find each of the following product:

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

Answer 5: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+nam×an=am+n. 

We have: 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 (x×x2)×(y2×y)×(z×z2) 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 ×(x1+2)×(y2+1)×(z1+2) 

= - 91/15 x3y3x3

Thus, the answer is = - 91/15 x3y3x3 

Question 6: Find each of the following product:

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

Answer 6: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+nam×an=am+n. 

We have: 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 ×(x3×x)×(z×z2)×y 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 ×(x3+1)×(z1+2)×y 

9/10 x4yz3=910x4yz3  

Thus, the answer is  9/10 x4yz3=910x4yz3  

Question 7: Find each of the following product:

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

Answer 7:To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+nam×an=am+n.
We have: 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 (a2×a3)×(b2×b2)×c2 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8  (a2+3)×(b2+2)×c2 

= - 1/6 a5b4c2 

Thus, the answer is = - 1/6 a5b4c2

Question 8: Find each of the following product: 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

Answer 8: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+nam×an=am+n.
We have: 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

=(7×1/4)×(x×x2)×(y×y)×z 

=(7×1/4)×(x1+2)×(y1+1)×z 

= 7/4 x3y2z

Thus, the answer is  7/4 x3y2z 

Question 9: Find each of the following product:
(7ab) × (−5ab2c) × (6abc2)

Answer 9: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e.,  am×an=am+nam×an=am+n.
We have: (7ab)×(5ab2c)×(6abc2)={7×(5)×6}×(a×a×a)×(b×b2×b)×(c×c2)={7×(5)×6}×(a1+1+1)×(b1+2+1)×(c1+2)=210a3b4c3 Thus, the answer is 210a3b4c3-210a3b4c3. 

Question 10: Find each of the following product:
(−5a) × (−10a2) × (−2a3)

Answer 10: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+nam×an=am+n.
We have:
(5a)×(10a2)×(2a3)={(5)×(10)×(2)}×(a×a2×a3)={(5)×(10)×(2)}×(a1+2+3)=100a6
Thus, the answer is 100a6-100a6. 

Question 11: Find each of the following product:
(−4x2) × (−6xy2) × (−3yz2)

Answer 11: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+nam×an=am+n.
We have:
(4x2)×(6xy2)×(3yz2)={(4)×(6)×(3)}×(x2×x)×(y2×y)×z2={(4)×(6)×(3)}×(x2+1)×(y2+1)×z2=72x3y3z2Thus, the answer is 72x3y3z2-72x3y3z2.

Question 12: Find each of the following product:

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

Answer 12: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+nam×an=am+n.
We have: 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 ×(a4×a2)×(b×b2) 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 ×a4+2×b1+2 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 ×a6×b3 

=− 3/5 a6b3 

Thus, the answer is  − 3/5 a6b3

Question 13: Find each of the following product:

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

Answer 13: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+nam×an=am+n.
We have:

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 ×(a×a×a2)×(b2×b)×(c2×c) 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 ×(a×a×a2)×(b2×b)×(c2×c) 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 ×(a1+1+2)×(b2+1)×(c2+1) 

=a4b3c3 

Thus, the answer is a4b3c3-a4b3c3. 

Question 14: Find each of the following product:

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

Answer 14: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e.,am×an=am+nam×an=am+n.
We have:

(4/3 u2vw)×(5uvw2)×(1/3 v2wu)  

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 ×(u2×u×u)×(v×v×v2)×(w×w2×w) 

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-2) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 ×(u2+1+1)×(v1+1+2)×(w1+2+1) 

= 20/9 u4v4w4 

Thus, the answer is  20/9 u4v4w4 

Question 15: Find each of the following product:
(0.5x)×(1/3 xy2z4)×(24x2yz)0.5x×13xy2z4×24x2yz

Answer 15: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+nam×an=am+n.
We have: 

(0.5x)×(1/3 xy2z4)×(24x2yz) 

=(0.5×1/3×24)×(x×x×x2)×(y2×y)×(z4×z) 

=(0.5×1/3×24)×(x1+1+2)×(y2+1)×(z4+1) 

=4x4y3z5 

Thus, the answer is 4x4y3z54x4y3z5. 

Question 16: Find each of the following product:

(4/3 pq2)×(1/4 p2r)×(16p2q2r2)43pq2×-14p2r×16p2q2r2 

Answer 16: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+nam×an=am+n.
We have: 

(4/3 pq2)×(1/4 p2r)×(16p2q2r2)

={4/3×(−1/4)×16}×(p×p2×p2)×(q2×q2)×(r×r2) 

={4/3×(− 1/4)×16}×(p1+2+2)×(q2+2)×(r1+2) 

= 16/3p5q4r3 

Thus, the answer is 1/3 p5q4r3-13p5q4r3. 

Question 17: Find each of the following product:
(2.3xy) × (0.1x) × (0.16)

Answer 17: To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+nam×an=am+n.
We have:
(2.3xy)×(0.1x)×(0.16)=(2.3×0.1×0.16)×(x×x)×y=(2.3×0.1×0.16)×(x1+1)×y=0.0368x2y
Thus, the answer is 0.0368x2y0.0368x2y.

Question 18: Express each of the following product as a monomials and verify the result in each case for x = 1:
(3x) × (4x) × (−5x)

Answer 18: We have to find the product of the expression in order to express it as a monomial.
To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+nam×an=am+n.
We have:
(3x)×(4x)×(5x)={3×4×(5)}×(x×x×x)={3×4×(5)}×(x1+1+1)=60x33x×4x×-5x=3×4×-5×x×x×x=3×4×-5×x1+1+1=-60x3
Substituting x = 1 in LHS, we get:
LHS =(3x)×(4x)×(5x)
=(3×1)×(4×1)×(−5×1)

=−60
Putting x = 1 in RHS, we get:
RHS =60x3=60(1)3=60×1=60 LHS = RHS for = 1; therefore, the result is correct
Thus, the answer is 60x3-60x3. 

Question 19: Express each of the following product as a monomials and verify the result in each case for x = 1:
(4x2) × (−3x) × (4/5 x3)

Answer 19: We have to find the product of the expression in order to express it as a monomial.
To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e., am×an=am+nam×an=am+n. 

We  have: 

(4x2)×(3x)×(4/5 x3) 

={4×(3)×4/5}×(x2×x×x3) 

={4×(3)×4/5}×(x2+1+3)

=− 48/5x6  

(4x2)×(3x)×(4/5 x3)=− 48/5x6 

Substituting x = 1 in LHS, we get: 

LHS=(4x2)×(3x)×(4/5 x3) 

=(4×12)×(3×1)×(4/5 ×13) 

=4×(3)×4/5

=48/5

Putting x = 1 in RHS, we get: 

RHS= 48/5 x6 
=48/5×16 

=48/5

 LHS = RHS for = 1; therefore, the result is correct
Thus, the answer is − 48/5 x6-485x6. 

Question 20: Express each of the following product as a monomials and verify the result in each case for x = 1:
(5x4) × (x2)3 × (2x)2

Answer 20: We have to find the product of the expression in order to express it as a monomial.

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e., am×an=am+n and (am)n=amn am×an=am+n and amn=amn .
We have:
(5x4)×(x2)3×(2x)2 =(5x4)×(x6)×(22×x2)=(5×22)×(x4×x6×x2)=(5×22)×(x4+6+2)=20x12
∴ (5x4)×(x2)3×(2x)2 =20x12 

Substituting x = 1 in LHS, we get:
LHS=(5x4)×(x2)3×(2