Open App

Class 8 Exam > Class 8 Notes > RD Sharma Solutions for Class 8 Mathematics > RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 )

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics PDF Download

Question 1: Multiply:
(5x + 3) by (7x + 2)

Answer 1: To multiply, we will use distributive law as follows:
$(5 x + 3) (7 x + 2) = 5 x (7 x + 2) + 3 (7 x + 2) = (5 x \times 7 x + 5 x \times 2) + (3 \times 7 x + 3 \times 2) = (35 x^{2} + 10 x) + (21 x + 6) = 35 x^{2} + 10 x + 21 x + 6 = 35 x^{2} + 31 x + 6$ Thus, the answer is $35 x^{2} + 31 x + 6$ .

Question 2: Multiply:
(2x + 8) by (x − 3)

Answer 2: To multiply the expressions, we will use the distributive law in the following way:
$(2 x + 8) (x - 3) = 2 x (x - 3) + 8 (x - 3) = (2 x \times x - 2 x \times 3) + (8 x - 8 \times 3) = (2 x^{2} - 6 x) + (8 x - 24) = 2 x^{2} - 6 x + 8 x - 24 = 2 x^{2} + 2 x - 24$ Thus, the answer is $2 x^{2} + 2 x - 24$ .

Question 3: Multiply:
(7x + y) by (x + 5y)

Answer 3: To multiply, we will use distributive law as follows:
$(7 x + y) (x + 5 y) = 7 x (x + 5 y) + y (x + 5 y) = 7 x^{2} + 35 x y + x y + 5 y^{2} = 7 x^{2} + 36 x y + 5 y^{2}$
Thus, the answer is $7 x^{2} + 36 x y + 5 y^{2}$ .

Question 4: Multiply:
(a − 1) by (0.1a² + 3)

Answer 4: To multiply, we will use distributive law as follows:
$(a - 1) (0.1 a^{2} + 3) = 0.1 a^{2} (a - 1) + 3 (a - 1) = 0.1 a^{3} - 0.1 a^{2} + 3 a - 3$
Thus, the answer is $0.1 a^{3} - 0.1 a^{2} + 3 a - 3$ .

Question 5: Multiply:
(3x² + y²) by (2x² + 3y²)

Answer 5: To multiply, we will use distributive law as follows:
$(3 x^{2} + y^{2}) (2 x^{2} + 3 y^{2}) = 3 x^{2} (2 x^{2} + 3 y^{2}) + y^{2} (2 x^{2} + 3 y^{2}) = 6 x^{4} + 9 x^{2} y^{2} + 2 x^{2} y^{2} + 3 y^{4} = 6 x^{4} + 11 x^{2} y^{2} + 3 y^{4}$ Thus, the answer is $6 x^{4} + 11 x^{2} y^{2} + 3 y^{4}$ .

Question 6: Multiply:

Answer 6: To multiply, we will use distributive law as follows:

Thus, the answer is

Question 7: Multiply:
(x⁶ − y⁶) by (x²+ y²)

Answer 7: To multiply, we will use distributive law as follows:

(x6−y6)(x2+y2)=x6(x2+y2)−y6(x2+y2)=(x8+x6y2)−(y6x2+y8)=x8+x6y2−y6x2−y8 Thus, the answer is $x^{8} + x^{6} y^{2} - y^{6} x^{2} - y^{8}$ .

Question 8: Multiply:
(x² + y²) by (3a + 2b)

Answer 8: To multiply, we will use distributive law as follows:

(x2+y2)(3a+2b)=x2(3a+2b)+y2(3a+2b)=3ax2+2bx2+3ay2+2by2 Thus, the answer is $3 a x^{2} + 2 b x^{2} + 3 a y^{2} + 2 b y^{2}$ .

Question 9: Multiply:
[−3d + (−7f)] by (5d + f)

Answer 9: To multiply, we will use distributive law as follows:
$[- 3 d + (- 7 f)] (5 d + f) = (- 3 d) (5 d + f) + (- 7 f) (5 d + f) = (- 15 d^{2} - 3 d f) + (- 35 d f - 7 f^{2}) = - 15 d^{2} - 3 d f - 35 d f - 7 f^{2} = - 15 d^{2} - 38 d f - 7 f^{2}$ Thus, the answer is $- 15 d^{2} - 38 d f - 7 f^{2}$ .

Question 10: Multiply:
(0.8a − 0.5b) by (1.5a − 3b)

Answer 10: To multiply, we will use distributive law as follows:
$(0.8 a - 0.5 b) (1.5 a - 3 b) = 0.8 a (1.5 a - 3 b) - 0.5 b (1.5 a - 3 b) = 1.2 a^{2} - 2.4 a b - 0.75 a b + 1.5 b^{2} = 1.2 a^{2} - 3.15 a b + 1.5 b^{2}$ Thus, the answer is $1.2 a^{2} - 3.15 a b + 1.5 b^{2}$ .

Question 11: Multiply:
(2x²y² − 5xy²) by (x² − y²)

Answer 11: To multiply, we will use distributive law as follows:

(2x2y2−5xy2)(x2−y2)=2x2y2(x2−y2)−5xy2(x2−y2)=2x4y2−2x2y4−5x3y2+5xy4 Thus, the answer is $2 x^{4} y^{2} - 2 x^{2} y^{4} - 5 x^{3} y^{2} + 5 x y^{4}$ .

Question 12: Multiply:

Answer 12: To multiply the expressions, we will use the distributive law in the following way:

Thus, the answer is

Question 13: Multiply:

Answer 13: To multiply, we will use distributive law as follows:

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

Thus, the answer is RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

Question 14: Multiply:

(3x²y − 5xy²) by

Answer 14: To multiply, we will use distributive law as follows:

(3x2y−5xy2) (1/5 x2+1/3 y2)

= 1/5 x2(3x2y−5xy2)+1/3 y2(3x2y−5xy2)

= 3/5 x4y−x3y2+x2y3−5/3 xy4

Thus, the answer is 3/5 x4y−x3y2+x2y3−5/3 xy4

Question 15: Multiply:
(2x² − 1) by (4x³ + 5x²)

Answer 15: To multiply, we will use distributive law as follows:
$(2 x^{2} - 1) (4 x^{3} + 5 x^{2}) = 2 x^{2} (4 x^{3} + 5 x^{2}) - 1 (4 x^{3} + 5 x^{2}) = 8 x^{5} + 10 x^{4} - 4 x^{3} - 5 x^{2}$ Thus, the answer is $8 x^{5} + 10 x^{4} - 4 x^{3} - 5 x^{2}$ .

Question 16: (2xy + 3y²) (3y² − 2)

Answer 16: To multiply, we will use distributive law as follows:
$(2 x y + 3 y^{2}) (3 y^{2} - 2) = 2 x y (3 y^{2} - 2) + 3 y^{2} (3 y^{2} - 2) = 6 x y^{3} - 4 x y + 9 y^{4} - 6 y^{2} = 9 y^{4} + 6 x y^{3} - 6 y^{2} - 4 x y$ Thus, the answer is $9 y^{4} + 6 x y^{3} - 6 y^{2} - 4 x y$ .

Question 17: Find the following product and verify the result for x = − 1, y = − 2:
(3x − 5y) (x + y)

Answer 17: To multiply, we will use distributive law as follows:
$(3 x - 5 y) (x + y) = 3 x (x + y) - 5 y (x + y) = 3 x^{2} + 3 x y - 5 x y - 5 y^{2} = 3 x^{2} - 2 x y - 5 y^{2}$ $∴$ $(3 x - 5 y) (x + y) = 3 x^{2} - 2 x y - 5 y^{2}$ .

Now, we put x = $-$ 1 and y = $-$ 2 on both sides to verify the result.

LHS=(3x−5y)(x+y)={3(−1)−5(−2)}{−1+(−2)}=(−3+10)(−3)=(7)(−3)=−21 RHS=3x2−2xy−5y2=3(−1)2−2(−1)(−2)−5(−2)2=3×1−4−5×4=3−4−20=−21 Because LHS is equal to RHS, the result is verified.
Thus, the answer is $3 x^{2} - 2 x y - 5 y^{2}$ .

Question 18: Find the following product and verify the result for x = − 1, y = − 2:
(x²y − 1) (3 − 2x²y)

Answer 18: To multiply, we will use distributive law as follows:

(x2y−1)(3−2x2y)=x2y(3−2x2y)−1×(3−2x2y)=3x2y−2x4y2−3+2x2y=5x2y−2x4y2−3 $∴$ $(x^{2} y - 1) (3 - 2 x^{2} y) = 5 x^{2} y - 2 x^{4} y^{2} - 3$

Now, we put x = $-$ 1 and y = $-$ 2 on both sides to verify the result.

LHS = (x2y−1)(3−2x2y)=[(−1)2(−2)−1][3−2(−1)2(−2)]=[1×(−2)−1][3−2×1×(−2)]=(−2−1)(3+4)=−3×7=−21RHS=5x2y−2x4y2−3=5(−1)2(−2)−2(−1)4(−2)2−3=[5×1×(−2)]−[2×1×4]−3=−10−8−3=−21 Because LHS is equal to RHS, the result is verified.
Thus, the answer is $5 x^{2} y - 2 x^{4} y^{2} - 3$ .

Question 19: Find the following product and verify the result for x = − 1, y = − 2:

Answer 19: To multiply, we will use distributive law as follows:

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

Now, we will put x = $-$ 1 and y = $-$ 2 on both the sides to verify the result.

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

=- 119/225

RHS= RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

= - 119/225

Because LHS is equal to RHS, the result is verified.

Thus, the answer is RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

Question 20: Simplify:
x²(x + 2y) (x − 3y)

Answer 20: To simplify, we will proceed as follows:

x2(x+2y)(x−3y)=[x2(x+2y)](x−3y)=(x3+2x2y)(x−3y)=x3(x−3y)+2x2y(x−3y)=x4−3x3y+2x3y−6x2y2=x4−x3y−6x2y2 Thus, the answer is $x^{4} - x^{3} y - 6 x^{2} y^{2}$ .

Question 21: Simplify:
(x² − 2y²) (x + 4y) x²y²

Answer 21: To simplify, we will proceed as follows:

(x2−2y2)(x+4y)x2y2=[x2(x+4y)−2y2(x+4y)]x2y2=(x3+4x2y−2xy2−8y3)x2y2=x5y2+4x4y3−2x3y4−8x2y5 Thus, the answer is $x^{5} y^{2} + 4 x^{4} y^{3} - 2 x^{3} y^{4} - 8 x^{2} y^{5}$ .

Question 22: Simplify:
a²b²(a + 2b)(3a + b)

Answer 22: To simplify, we will proceed as follows:

a2b2(a+2b)(3a+b)=[a2b2(a+2b)](3a+b)=(a3b2+2a2b3)(3a+b)=3a(a3b2+2a2b3)+b(a3b2+2a2b3)=3a4b2+6a3b3+a3b3+2a2b4=3a4b2+7a3b3+2a2b4 Thus, the answer is $3 a^{4} b^{2} + 7 a^{3} b^{3} + 2 a^{2} b^{4}$ .

Question 23: Simplify:
x²(x − y) y²(x + 2y)

Answer 23: To simplify, we will proceed as follows:

x2(x−y)y2(x+2y)=[x2(x−y)][y2(x+2y)]=(x3−x2y)(xy2+2y3)=x3(xy2+2y3)−x2y(xy2+2y3)=x4y2+2x3y3−[x3y3+2x2y4]=x4y2+2x3y3−x3y3−2x2y4=x4y2+x3y3−2x2y4 Thus, the answer is $x^{4} y^{2} + x^{3} y^{3} - 2 x^{2} y^{4}$ .

Question 24: Simplify:
(x³ − 2x² + 5x − 7)(2x − 3)

Answer 24: To simplify, we will proceed as follows:

(x3−2x2+5x−7)(2x−3)=2x(x3−2x2+5x−7)−3(x3−2x2+5x−7)=2x4−4x3+10x2−14x−3x3+6x2−15x+21 $= 2 x^{4} - 4 x^{3} - 3 x^{3} + 10 x^{2} + 6 x^{2} - 14 x - 15 x + 21$ (Rearranging)
$= 2 x^{4} - 7 x^{3} + 16 x^{2} - 29 x + 21$ (Combining like terms)

Thus, the answer is $2 x^{4} - 7 x^{3} + 16 x^{2} - 29 x + 21$ .

Question 25: Simplify:
(5x + 3)(x − 1)(3x − 2)

Answer 25: To simplify, we will proceed as follows:
$(5 x + 3) (x - 1) (3 x - 2) = [(5 x + 3) (x - 1)] (3 x - 2)$ $= [5 x (x - 1) + 3 (x - 1)] (3 x - 2)$ (Distributive law)
$= [5 x^{2} - 5 x + 3 x - 3] (3 x - 2) = [5 x^{2} - 2 x - 3] (3 x - 2) = 3 x (5 x^{2} - 2 x - 3) - 2 (5 x^{2} - 2 x - 3) = 15 x^{3} - 6 x^{2} - 9 x - [10 x^{2} - 4 x - 6] = 15 x^{3} - 6 x^{2} - 9 x - 10 x^{2} + 4 x + 6$ $= 15 x^{3} - 6 x^{2} - 10 x^{2} - 9 x + 4 x + 6$ (Rearranging)
$= 15 x^{3} - 16 x^{2} - 5 x + 6$ (Combining like terms)

Thus, the answer is $15 x^{3} - 16 x^{2} - 5 x + 6$ .

Question 26: Simplify:
(5 − x)(6 − 5x)( 2 − x)

Answer 26: To simplify, we will proceed as follows:

(5−x)(6−5x)(2−x)=[(5−x)(6−5x)](2−x) $= [5 (6 - 5 x) - x (6 - 5 x)] (2 - x)$ (Distributive law)
$= (30 - 25 x - 6 x + 5 x^{2}) (2 - x) = (30 - 31 x + 5 x^{2}) (2 - x) = 2 (30 - 31 x + 5 x^{2}) - x (30 - 31 x + 5 x^{2}) = 60 - 62 x + 10 x^{2} - 30 x + 31 x^{2} - 5 x^{3}$ $= 60 - 62 x - 30 x + 10 x^{2} + 31 x^{2} - 5 x^{3}$ (Rearranging)
$= 60 - 92 x + 41 x^{2} - 5 x^{3}$ (Combining like terms)

Thus, the answer is $60 - 92 x + 41 x^{2} - 5 x^{3}$ .

Question 27: Simplify:
(2x² + 3x − 5)(3x² − 5x + 4)

Answer 27: To simplify, we will proceed as follows:

$(2 x^{2} + 3 x - 5) (3 x^{2} - 5 x + 4)$
$= 2 x^{2} (3 x^{2} - 5 x + 4) + 3 x (3 x^{2} - 5 x + 4) - 5 (3 x^{2} - 5 x + 4)$ (Distributive law)
$= 6 x^{4} - 10 x^{3} + 8 x^{2} + 9 x^{3} - 15 x^{2} + 12 x - 15 x^{2} + 25 x - 20$
$= 6 x^{4} - 10 x^{3} + 9 x^{3} + 8 x^{2} - 15 x^{2} - 15 x^{2} + 12 x + 25 x - 20$ (Rearranging)
$= 6 x^{4} - x^{3} - 22 x^{2} + 36 x - 20$ (Combining like terms)

Thus, the answer is $6 x^{4} - x^{3} - 22 x^{2} + 36 x - 20$ .

Question 28: Simplify:
(3x − 2)(2x − 3) + (5x − 3)(x + 1)

Answer 28: To simplify, we will proceed as follows:

(3x−2)(2x−3)+(5x−3)(x+1)=[(3x−2)(2x−3)]+[(5x−3)(x+1)] $= [3 x (2 x - 3) - 2 (2 x - 3)] + [5 x (x + 1) - 3 (x + 1)]$ (Distributive law)
$= 6 x^{2} - 9 x - 4 x + 6 + 5 x^{2} + 5 x - 3 x - 3$
$= 6 x^{2} + 5 x^{2} - 9 x - 4 x + 5 x - 3 x - 3 + 6$ (Rearranging)
$= 11 x^{2} - 11 x + 3$ (Combining like terms)

Thus, the answer is $11 x^{2} - 11 x + 3$ .

Question 29: Simplify:
(5x − 3)(x + 2) − (2x + 5)(4x − 3)

Answer 29: To simplify, we will proceed as follows:

(5x−3)(x+2)−(2x+5)(4x−3)=[(5x−3)(x+2)]−[(2x+5)(4x−3)] $= [5 x (x + 2) - 3 (x + 2)] - [2 x (4 x - 3) + 5 (4 x - 3)]$ (Distributive law)
$= 5 x^{2} + 10 x - 3 x - 6 - 8 x^{2} + 6 x - 20 x + 15$
$= 5 x^{2} - 8 x^{2} + 10 x - 3 x + 6 x - 20 x - 6 + 15$ (Rearranging)
$= 5 x^{2} - 8 x^{2} + 10 x - 3 x + 6 x - 20 x - 6 + 15 = - 3 x^{2} - 7 x + 9$ (Combining like terms)

Hence, the answer is $- 3 x^{2} - 7 x + 9$ .

Question 30: Simplify:
(3x + 2y)(4x + 3y) − (2x − y)(7x − 3y)

Answer 30: To simplify, we will proceed as follows:

(3x+2y)(4x+3y)−(2x−y)(7x−3y)=[(3x+2y)(4x+3y)]−[(2x−y)(7x−3y)] $= [3 x (4 x + 3 y) + 2 y (4 x + 3 y)] - [2 x (7 x - 3 y) - y (7 x - 3 y)]$ (Distributive law)
$= 12 x^{2} + 9 x y + 8 x y + 6 y^{2} - [14 x^{2} - 6 x y - 7 x y + 3 y^{2}] = 12 x^{2} + 9 x y + 8 x y + 6 y^{2} - 14 x^{2} + 6 x y + 7 x y - 3 y^{2}$

$= 12 x^{2} + 9 x y + 8 x y + 6 y^{2} - [14 x^{2} - 6 x y - 7 x y + 3 y^{2}] = 12 x^{2} + 9 x y + 8 x y + 6 y^{2} - 14 x^{2} + 6 x y + 7 x y - 3 y^{2}$

Question 31: Simplify:
(x² − 3x + 2)(5x − 2) − (3x² + 4x − 5)(2x − 1)

Answer 31: To simplify, we will proceed as follows:

$(x^{2} - 3 x + 2) (5 x - 2) - (3 x^{2} + 4 x - 5) (2 x - 1) = [(x^{2} - 3 x + 2) (5 x - 2)] - [(3 x^{2} + 4 x - 5) (2 x - 1)]$
$= [5 x (x^{2} - 3 x + 2) - 2 (x^{2} - 3 x + 2)] - [2 x (3 x^{2} + 4 x - 5) - 1 \times (3 x^{2} + 4 x - 5)]$ (Distributive law)
$= [5 x^{3} - 15 x^{2} + 10 x - (2 x^{2} - 6 x + 4)] - [6 x^{3} + 8 x^{2} - 10 x - 3 x^{2} - 4 x + 5] = [5 x^{3} - 15 x^{2} + 10 x - 2 x^{2} + 6 x - 4] - [6 x^{3} + 8 x^{2} - 10 x - 3 x^{2} - 4 x + 5] = 5 x^{3} - 15 x^{2} + 10 x - 2 x^{2} + 6 x - 4 - 6 x^{3} - 8 x^{2} + 10 x + 3 x^{2} + 4 x - 5$
$= 5 x^{3} - 6 x^{3} - 15 x^{2} - 2 x^{2} - 8 x^{2} + 3 x^{2} + 10 x + 6 x + 10 x + 4 x - 5 - 4$ (Rearranging)
$= - x^{3} - 22 x^{2} + 30 x - 9$ (Combining like terms)

Thus, the answer is $- x^{3} - 22 x^{2} + 30 x - 9$ .

Question 32: Simplify:
(x³ − 2x² + 3x − 4) (x −1) − (2x − 3)(x² − x + 1)

Answer 32: To simplify,we will proceed as follows:

(x3−2x2+3x−4)(x−1)−(2x−3)(x2−x+1)=[(x3−2x2+3x−4)(x−1)]−[(2x−3)(x2−x+1)] $= [x (x^{3} - 2 x^{2} + 3 x - 4) - 1 (x^{3} - 2 x^{2} + 3 x - 4)] - [2 x (x^{2} - x + 1) - 3 (x^{2} - x + 1)]$ (Distributive law)

=[x(x3−2x2+3x−4)−1(x3−2x2+3x−4)]−[2x(x2−x+1)−3(x2−x+1)]=x4−2x3+3x2−4x−x3+2x2−3x+4−[2x3−2x2+2x−3x2+3x−3]=x4−2x3+3x2−4x−x3+2x2−3x+4−2x3+2x2−2x+3x2−3x+3 $= x^{4} - 2 x^{3} - 2 x^{3} - x^{3} + 3 x^{2} + 2 x^{2} + 2 x^{2} + 3 x^{2} - 4 x - 3 x - 2 x - 3 x + 4 + 3$ (Rearranging)
$= x^{4} - 5 x^{3} + 10 x^{2} - 12 x + 7$ (Combining like terms)

Thus, the answer is $x^{4} - 5 x^{3} + 10 x^{2} - 12 x + 7$ .

The document RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.

All you need of Class 8 at this link: Class 8

	RD Sharma Solutions for Class 8 Mathematics 88 docs

RD Sharma Solutions for Class 8 Mathematics

88 docs

Join Course for Free

Top Courses for Class 8

View all

Related Exams

Class 8

About this Document

	4.96/5 Rating
	Nov 22, 2024 Last updated

Document Description: RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) for Class 8 2024 is part of RD Sharma Solutions for Class 8 Mathematics preparation. The notes and questions for RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) have been prepared according to the Class 8 exam syllabus. Information about RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) covers topics like and RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Example, for Class 8 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ).

Introduction of RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) in English is available as part of our RD Sharma Solutions for Class 8 Mathematics for Class 8 & RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) in Hindi for RD Sharma Solutions for Class 8 Mathematics course. Download more important topics related with notes, lectures and mock test series for Class 8 Exam by signing up for free. Class 8: RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

Description

Full syllabus notes, lecture & questions for RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics - Class 8 | Plus excerises question with solution to help you revise complete syllabus for RD Sharma Solutions for Class 8 Mathematics | Best notes, free PDF download

Information about RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 )

In this doc you can find the meaning of RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) defined & explained in the simplest way possible. Besides explaining types of RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) theory, EduRev gives you an ample number of questions to practice RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) tests, examples and also practice Class 8 tests

	RD Sharma Solutions for Class 8 Mathematics 88 docs

RD Sharma Solutions for Class 8 Mathematics

88 docs

Join Course for Free

Download as PDF

Explore Courses for Class 8 exam

Top Courses for Class 8

Explore Courses

Signup for Free!

Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.

Start learning for Free

10M+ students study on EduRev

ppt

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

Free

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

MCQs

Semester Notes

Exam

video lectures

Summary

shortcuts and tricks

Previous Year Questions with Solutions

Important questions

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

study material

practice quizzes

Objective type Questions

past year papers

pdf

Viva Questions

Sample Paper

mock tests for examination

Extra Questions

;

Additional Information about RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) for Class 8 Preparation

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Free PDF Download

The RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) is an invaluable resource that delves deep into the core of the Class 8 exam. These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective. With the help of these notes, you can grasp complex subjects quickly, revise important points easily, and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner, allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material, or toppers' notes, this PDF has got you covered. Download the RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) now and kickstart your journey towards success in the Class 8 exam.

Importance of RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 )

The importance of RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) cannot be overstated, especially for Class 8 aspirants. This document holds the key to success in the Class 8 exam. It offers a detailed understanding of the concept, providing invaluable insights into the topic. By knowing the concepts well in advance, students can plan their preparation effectively. Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes offer in-depth insights into the specific topic to help you master it with ease. This comprehensive document covers all aspects related to RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ). It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation. Practice papers and question papers enable you to assess your progress effectively. Additionally, the paper analysis provides valuable tips for tackling the exam strategically. Access to Toppers' notes gives you an edge in understanding complex concepts. Whether you're a beginner or aiming for advanced proficiency, RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Notes on EduRev are your ultimate resource for success.

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Class 8 Questions

The "RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) Class 8 Questions" guide is a valuable resource for all aspiring students preparing for the Class 8 exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

Study RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) on the App

Students of Class 8 can study RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ), students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc. The EduRev App will make your learning easier as you can access it from anywhere you want. The content of RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) is prepared as per the latest Class 8 syllabus.

Education Revolution

Signup to see your scores go up within 7 days!

Access 1000+ FREE Docs, Videos and Tests

Continue with Google

Takes less than 10 seconds to signup

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics PDF Download

Question 1: Multiply:(5x + 3) by (7x + 2)

Answer 1: To multiply, we will use distributive law as follows:(5x+3)(7x+2)=5x(7x+2)+3(7x+2)=(5x×7x+5x×2)+(3×7x+3×2)=(35x2+10x)+(21x+6)=35x2+10x+21x+6=35x2+31x+6Thus, the answer is 35x2+31x+635x2+31x+6.

Question 2: Multiply:(2x + 8) by (x − 3)

Question 3: Multiply:(7x + y) by (x + 5y)

Answer 3: To multiply, we will use distributive law as follows:(7x+y)(x+5y)=7x(x+5y)+y(x+5y)=7x2+35xy+xy+5y2=7x2+36xy+5y27x+yx+5y=7xx+5y+yx+5y=7x2+35xy+xy+5y2=7x2+36xy+5y2Thus, the answer is 7x2+36xy+5y27x2+36xy+5y2.

Question 4: Multiply:(a − 1) by (0.1a2 + 3)

Answer 4: To multiply, we will use distributive law as follows:(a−1)(0.1a2+3)=0.1a2(a−1)+3(a−1)=0.1a3−0.1a2+3a−3a-10.1a2+3=0.1a2a-1+3a-1=0.1a3-0.1a2+3a-3Thus, the answer is 0.1a3−0.1a2+3a−30.1a3-0.1a2+3a-3.

Question 5: Multiply:(3x2 + y2) by (2x2 + 3y2)

Answer 5: To multiply, we will use distributive law as follows:(3x2+y2)(2x2+3y2)=3x2(2x2+3y2)+y2(2x2+3y2)=6x4+9x2y2+2x2y2+3y4=6x4+11x2y2+3y43x2+y22x2+3y2=3x22x2+3y2+y22x2+3y2=6x4+9x2y2+2x2y2+3y4=6x4+11x2y2+3yThus, the answer is 6x4+11x2y2+3y46x4+11x2y2+3y4.

Question 6: Multiply:

Answer 6: To multiply, we will use distributive law as follows:

Question 7: Multiply:(x6 − y6) by (x2 + y2)

Answer 7: To multiply, we will use distributive law as follows:

Question 8: Multiply:(x2 + y2) by (3a + 2b)

Answer 8: To multiply, we will use distributive law as follows:

Question 9: Multiply:[−3d + (−7f)] by (5d + f)

Answer 9: To multiply, we will use distributive law as follows:[−3d+(−7f)](5d+f)=(−3d)(5d+f)+(−7f)(5d+f)=(−15d2−3df)+(−35df−7f2)=−15d2−3df−35df−7f2=−15d2−38df−7f2Thus, the answer is −15d2−38df−7f2-15d2-38df-7f2.

Question 10: Multiply:(0.8a − 0.5b) by (1.5a − 3b)

Answer 10: To multiply, we will use distributive law as follows:(0.8a−0.5b)(1.5a−3b)=0.8a(1.5a−3b)−0.5b(1.5a−3b)=1.2a2−2.4ab−0.75ab+1.5b2=1.2a2−3.15ab+1.5b2Thus, the answer is 1.2a2−3.15ab+1.5b21.2a2-3.15ab+1.5b2.

Question 11: Multiply:(2x2y2 − 5xy2) by (x2 − y2)

Answer 11: To multiply, we will use distributive law as follows:

(2x2y2−5xy2)(x2−y2)=2x2y2(x2−y2)−5xy2(x2−y2)=2x4y2−2x2y4−5x3y2+5xy4 Thus, the answer is 2x4y2−2x2y4−5x3y2+5xy42x4y2-2x2y4-5x3y2+5xy4.

Question 12: Multiply:

Answer 12: To multiply the expressions, we will use the distributive law in the following way:

Question 13: Multiply:

Answer 13: To multiply, we will use distributive law as follows:

Question 14: Multiply:

Answer 14: To multiply, we will use distributive law as follows:

Question 15: Multiply:(2x2 − 1) by (4x3 + 5x2)

Answer 15: To multiply, we will use distributive law as follows:(2x2−1)(4x3+5x2)=2x2(4x3+5x2)−1(4x3+5x2)=8x5+10x4−4x3−5x2Thus, the answer is 8x5+10x4−4x3−5x28x5+10x4-4x3-5x2.

Question 16: (2xy + 3y2) (3y2 − 2)

Answer 16: To multiply, we will use distributive law as follows:(2xy+3y2)(3y2−2)=2xy(3y2−2)+3y2(3y2−2)=6xy3−4xy+9y4−6y2=9y4+6xy3−6y2−4xyThus, the answer is 9y4+6xy3−6y2−4xy9y4+6xy3-6y2-4xy.

Question 17: Find the following product and verify the result for x = − 1, y = − 2:(3x − 5y) (x + y)

Answer 17: To multiply, we will use distributive law as follows:(3x−5y)(x+y)=3x(x+y)−5y(x+y)=3x2+3xy−5xy−5y2=3x2−2xy−5y2∴∴ (3x−5y)(x+y)=3x2−2xy−5y23x-5yx+y=3x2-2xy-5y2.

Now, we put x = −-1 and y = −-2 on both sides to verify the result.

LHS=(3x−5y)(x+y)={3(−1)−5(−2)}{−1+(−2)}=(−3+10)(−3)=(7)(−3)=−21 RHS=3x2−2xy−5y2=3(−1)2−2(−1)(−2)−5(−2)2=3×1−4−5×4=3−4−20=−21 Because LHS is equal to RHS, the result is verified.Thus, the answer is 3x2−2xy−5y23x2-2xy-5y2.

Question 18: Find the following product and verify the result for x = − 1, y = − 2:(x2y − 1) (3 − 2x2y)

Answer 18: To multiply, we will use distributive law as follows:

Question 19: Find the following product and verify the result for x = − 1, y = − 2:

Answer 19: To multiply, we will use distributive law as follows:

Question 20: Simplify:x2(x + 2y) (x − 3y)

Answer 20: To simplify, we will proceed as follows:

Question 21: Simplify:(x2 − 2y2) (x + 4y) x2y2

Answer 21: To simplify, we will proceed as follows:

Question 22: Simplify:a2b2(a + 2b)(3a + b)

Answer 22: To simplify, we will proceed as follows:

Question 23: Simplify:x2(x − y) y2(x + 2y)

Answer 23: To simplify, we will proceed as follows:

Question 24: Simplify:(x3 − 2x2 + 5x − 7)(2x − 3)

Answer 24: To simplify, we will proceed as follows:

Question 25: Simplify:(5x + 3)(x − 1)(3x − 2)

Question 26: Simplify:(5 − x)(6 − 5x)( 2 − x)

Answer 26: To simplify, we will proceed as follows:

Question 27: Simplify:(2x2 + 3x − 5)(3x2 − 5x + 4)

Answer 27: To simplify, we will proceed as follows:

Question 28: Simplify:(3x − 2)(2x − 3) + (5x − 3)(x + 1)

Answer 28: To simplify, we will proceed as follows:

Question 29: Simplify:(5x − 3)(x + 2) − (2x + 5)(4x − 3)

Answer 29: To simplify, we will proceed as follows:

Question 30: Simplify:(3x + 2y)(4x + 3y) − (2x − y)(7x − 3y)

Answer 30: To simplify, we will proceed as follows:

Question 31: Simplify:(x2 − 3x + 2)(5x − 2) − (3x2 + 4x − 5)(2x − 1)

Answer 31: To simplify, we will proceed as follows:

Question 32: Simplify:(x3 − 2x2 + 3x − 4) (x −1) − (2x − 3)(x2 − x + 1)

Answer 32: To simplify,we will proceed as follows:

RD Sharma Solutions for Class 8 Mathematics

Top Courses for Class 8

RD Sharma Solutions for Class 8 Mathematics

Top Courses for Class 8

ppt

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

Free

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-4 ) | RD Sharma Solutions for Class 8 Mathematics

MCQs

Semester Notes

Exam

video lectures

Summary

shortcuts and tricks

Question 1: Multiply:
(5x + 3) by (7x + 2)

Question 2: Multiply:
(2x + 8) by (x − 3)

Question 3: Multiply:
(7x + y) by (x + 5y)

Answer 3: To multiply, we will use distributive law as follows:
$(7 x + y) (x + 5 y) = 7 x (x + 5 y) + y (x + 5 y) = 7 x^{2} + 35 x y + x y + 5 y^{2} = 7 x^{2} + 36 x y + 5 y^{2}$
Thus, the answer is $7 x^{2} + 36 x y + 5 y^{2}$ .

Question 4: Multiply:
(a − 1) by (0.1a² + 3)

Answer 4: To multiply, we will use distributive law as follows:
$(a - 1) (0.1 a^{2} + 3) = 0.1 a^{2} (a - 1) + 3 (a - 1) = 0.1 a^{3} - 0.1 a^{2} + 3 a - 3$
Thus, the answer is $0.1 a^{3} - 0.1 a^{2} + 3 a - 3$ .

Question 5: Multiply:
(3x² + y²) by (2x² + 3y²)

Question 7: Multiply:
(x⁶ − y⁶) by (x²+ y²)

Question 8: Multiply:
(x² + y²) by (3a + 2b)

Question 9: Multiply:
[−3d + (−7f)] by (5d + f)

Question 10: Multiply:
(0.8a − 0.5b) by (1.5a − 3b)

Answer 10: To multiply, we will use distributive law as follows:
$(0.8 a - 0.5 b) (1.5 a - 3 b) = 0.8 a (1.5 a - 3 b) - 0.5 b (1.5 a - 3 b) = 1.2 a^{2} - 2.4 a b - 0.75 a b + 1.5 b^{2} = 1.2 a^{2} - 3.15 a b + 1.5 b^{2}$ Thus, the answer is $1.2 a^{2} - 3.15 a b + 1.5 b^{2}$ .

Question 11: Multiply:
(2x²y² − 5xy²) by (x² − y²)

(2x2y2−5xy2)(x2−y2)=2x2y2(x2−y2)−5xy2(x2−y2)=2x4y2−2x2y4−5x3y2+5xy4 Thus, the answer is $2 x^{4} y^{2} - 2 x^{2} y^{4} - 5 x^{3} y^{2} + 5 x y^{4}$ .

Question 15: Multiply:
(2x² − 1) by (4x³ + 5x²)

Answer 15: To multiply, we will use distributive law as follows:
$(2 x^{2} - 1) (4 x^{3} + 5 x^{2}) = 2 x^{2} (4 x^{3} + 5 x^{2}) - 1 (4 x^{3} + 5 x^{2}) = 8 x^{5} + 10 x^{4} - 4 x^{3} - 5 x^{2}$ Thus, the answer is $8 x^{5} + 10 x^{4} - 4 x^{3} - 5 x^{2}$ .

Question 16: (2xy + 3y²) (3y² − 2)

Answer 16: To multiply, we will use distributive law as follows:
$(2 x y + 3 y^{2}) (3 y^{2} - 2) = 2 x y (3 y^{2} - 2) + 3 y^{2} (3 y^{2} - 2) = 6 x y^{3} - 4 x y + 9 y^{4} - 6 y^{2} = 9 y^{4} + 6 x y^{3} - 6 y^{2} - 4 x y$ Thus, the answer is $9 y^{4} + 6 x y^{3} - 6 y^{2} - 4 x y$ .

Question 17: Find the following product and verify the result for x = − 1, y = − 2:
(3x − 5y) (x + y)

Answer 17: To multiply, we will use distributive law as follows:
$(3 x - 5 y) (x + y) = 3 x (x + y) - 5 y (x + y) = 3 x^{2} + 3 x y - 5 x y - 5 y^{2} = 3 x^{2} - 2 x y - 5 y^{2}$ $∴$ $(3 x - 5 y) (x + y) = 3 x^{2} - 2 x y - 5 y^{2}$ .

Now, we put x = $-$ 1 and y = $-$ 2 on both sides to verify the result.

LHS=(3x−5y)(x+y)={3(−1)−5(−2)}{−1+(−2)}=(−3+10)(−3)=(7)(−3)=−21 RHS=3x2−2xy−5y2=3(−1)2−2(−1)(−2)−5(−2)2=3×1−4−5×4=3−4−20=−21 Because LHS is equal to RHS, the result is verified.
Thus, the answer is $3 x^{2} - 2 x y - 5 y^{2}$ .

Question 18: Find the following product and verify the result for x = − 1, y = − 2:
(x²y − 1) (3 − 2x²y)

Question 20: Simplify:
x²(x + 2y) (x − 3y)

Question 21: Simplify:
(x² − 2y²) (x + 4y) x²y²

Question 22: Simplify:
a²b²(a + 2b)(3a + b)

Question 23: Simplify:
x²(x − y) y²(x + 2y)

Question 24: Simplify:
(x³ − 2x² + 5x − 7)(2x − 3)

Question 25: Simplify:
(5x + 3)(x − 1)(3x − 2)

Question 26: Simplify:
(5 − x)(6 − 5x)( 2 − x)

Question 27: Simplify:
(2x² + 3x − 5)(3x² − 5x + 4)

Question 28: Simplify:
(3x − 2)(2x − 3) + (5x − 3)(x + 1)

Question 29: Simplify:
(5x − 3)(x + 2) − (2x + 5)(4x − 3)

Question 30: Simplify:
(3x + 2y)(4x + 3y) − (2x − y)(7x − 3y)

Question 31: Simplify:
(x² − 3x + 2)(5x − 2) − (3x² + 4x − 5)(2x − 1)

Question 32: Simplify:
(x³ − 2x² + 3x − 4) (x −1) − (2x − 3)(x² − x + 1)