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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-3)

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) | RD Sharma Solutions for Class 8 Mathematics PDF Download

Question 1: Find the following product:
2a³(3a + 5b)

Answer 1: To find the product, we will use distributive law as follows:
$2 a^{3} (3 a + 5 b) = 2 a^{3} \times 3 a + 2 a^{3} \times 5 b = (2 \times 3) (a^{3} \times a) + (2 \times 5) a^{3} b = (2 \times 3) a^{3 + 1} + (2 \times 5) a^{3} b = 6 a^{4} + 10 a^{3} b$
Thus, the answer is $6 a^{4} + 10 a^{3} b$ .

Question 2: Find the following product:
−11a(3a + 2b)

Answer 2: To find the product, we will use distributive law as follows:
$- 11 a (3 a + 2 b) = (- 11 a) \times 3 a + (- 11 a) \times 2 b = (- 11 \times 3) \times (a \times a) + (- 11 \times 2) \times (a \times b) = (- 33) \times (a^{1 + 1}) + (- 22) \times (a \times b) = - 33 a^{2} - 22 a b$
Thus, the answer is $- 33 a^{2} - 22 a b$ .

Question 3: Find the following product:
−5a(7a − 2b)

Answer 3: To find the product, we will use distributive law as follows:
$- 5 a (7 a - 2 b) = (- 5 a) \times 7 a + (- 5 a) \times (- 2 b) = (- 5 \times 7) \times (a \times a) + (- 5 \times (- 2)) \times (a \times b) = (- 35) \times (a^{1 + 1}) + (10) \times (a \times b) = - 35 a^{2} + 10 a b$
Thus, the answer is $- 35 a^{2} + 10 a b$ .

Question 4: Find the following product:
−11y²(3y + 7)

Answer 4: To find the product, we will use distributive law as follows:

$- 11 y^{2} (3 y + 7) = (- 11 y^{2}) \times 3 y + (- 11 y^{2}) \times 7 = (- 11 \times 3) (y^{2} \times y) + (- 11 \times 7) \times (y^{2}) = (- 33) (y^{2 + 1}) + (- 77) \times (y^{2}) = - 33 y^{3} - 77 y^{2}$ =−33y³−77y²

Thus, the answer is −33y³−77y².

Question 5: Find the following product:
$\frac{6 x}{5} (x^{3} + y^{3})$

Answer 5: To find the product, we will use distributive law as follows:

6x/5 (x3+y3)

=6x/5×x3+6x/5×y3

=6/5×(x×x3)+6/5×(x×y3)

=6/5×(x1+3)+6/5×(x×y3)

Thus, the answer is .

Question 6: xy(x³ − y³)

Answer 6: To find the product, we will use the distributive law in the following way:
$x y (x^{3} - y^{3}) = x y \times x^{3} - x y \times y^{3} = (x \times x^{3}) \times y - x \times (y \times y^{3}) = x^{1 + 3} y - x y^{1 + 3} = x^{4} y - x y^{4}$
Thus, the answer is $x^{4} y - x y^{4}$ .

Question 7: Find the following product:
0.1y(0.1x⁵ + 0.1y)

Answer 7: To find the product, we will use distributive law as follows:
$0.1 y (0.1 x^{5} + 0.1 y) = (0.1 y) (0.1 x^{5}) + (0.1 y) (0.1 y) = (0.1 \times 0.1) (y \times x^{5}) + (0.1 \times 0.1) (y \times y) = (0.1 \times 0.1) (x^{5} \times y) + (0.1 \times 0.1) (y^{1 + 1}) = 0.01 x^{5} y + 0.01 y^{2}$
Thus, the answer is $0.01 x^{5} y + 0.01 y^{2}$ .

Question 8: Find the following product:

Answer 8: To find the product, we will use distributive law as follows:

={{−7/4×(−50)}(a×a²)×(b²×b²)×(c×c²)}−{(625)(−50)(a²×a²)×(b²)×(c²×c²)}

={−7/4×(−50)}(a1+2b2+2c1+2)−{(6/25)(−50)(a2+2b2c2+2)}

=175/2 a3b4c3−(−12a4b2c4)

=175/2 a3b4c3+12a4b2c4

Thus, the answer is 175/2 $\frac{175}{2} a^{3} b^{4} c^{3} + 12 a^{4} b^{2} c^{4}$ .

Question 9: Find the following product:

Answer 9: To find the product, we will use the distributive law in the following way:

(x×x)×(y×y)×(z×z2)}−
(x×x)×(y×y2)×(z×z3)}

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) | RD Sharma Solutions for Class 8 Mathematics (x¹⁺¹y¹⁺¹z¹⁺²)}− (x¹⁺¹y¹⁺²z¹⁺³)}

$- 11 y^{2} (3 y + 7) = (- 11 y^{2}) \times 3 y + (- 11 y^{2}) \times 7 = (- 11 \times 3) (y^{2} \times y) + (- 11 \times 7) \times (y^{2}) = (- 33) (y^{2 + 1}) + (- 77) \times (y^{2}) = - 33 y^{3} - 77 y^{2}$

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) | RD Sharma Solutions for Class 8 Mathematics -

=−4/9 x2y2z3+2/3 x2y3z4

Thus, the answer is −4/9 x2y2z3+2/3 x2y3z4

Question 10: Find the following product:

Answer 10: To find the product, we will use distributive law as follows:

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) | RD Sharma Solutions for Class 8 Mathematics

=−2/3 x3y2z2+1/9 x2y2z3

Thus, the answer is −2/3 x3y2z2+1/9 x2y2z3

Question 11: Find the following product:
1.5x(10x²y − 100xy²)

Answer 11: To find the product, we will use distributive law as follows:
$1.5 x (10 x^{2} y - 100 x y^{2}) = (1.5 x \times 10 x^{2} y) - (1.5 x \times 100 x y^{2}) = (15 x^{1 + 2} y) - (150 x^{1 + 1} y^{2}) = 15 x^{3} y - 150 x^{2} y^{2}$ Thus, the answer is $15 x^{3} y - 150 x^{2} y^{2}$ .

Question 12: Find the following product:
4.1xy(1.1x − y)

Answer 12: To find the product, we will use distributive law as follows:
$4.1 x y (1.1 x - y) = (4.1 x y \times 1.1 x) - (4.1 x y \times y) = \{(4.1 \times 1.1) \times x y \times x\} - (4.1 x y \times y) = (4.51 x^{1 + 1} y) - (4.1 x y^{1 + 1}) = 4.51 x^{2} y - 4.1 x y^{2}$ Thus, the answer is $4.51 x^{2} y - 4.1 x y^{2}$ .

Question 13: Find the following product:

250.5xy $(x z + \frac{y}{10})$

Answer 13: To find the product, we will use distributive law as follows:

$250.5 x y (x z + \frac{y}{10}) = 250.5 x y \times x z + 250.5 x y \times \frac{y}{10} = 250.5 x^{1 + 1} y z + 25.05 x y^{1 + 1} = 250.5 x^{2} y z + 25.05 x y^{2}$ Thus, the answer is $250.5 x^{2} y z + 25.05 x y^{2}$ .

Question 14: Find the following product:

Answer 14: To find the product, we will use distributive law as follows:

Thus, the answer is

Question 15: Find the following product:

4/3 a(a2 + b2 − 3c2)43a(a2 + b2 - 3c2)

Answer 15: To find the product, we will use distributive law as follows:

4/3 a(a2+b2−3c2)=4/3 a×a2+4/3 a×b2−4/3 a×3c2=4/3 a1+2+4/3 ab2−4ac2=4/3 a3+4/3 ab2−4ac2

Thus, the answer is 4/3 a3+4/3 ab2−4ac2

Question 16: Find the product 24x² (1 − 2x) and evaluate its value for x = 3.

Answer 16: To find the product, we will use distributive law as follows:
$24 x^{2} (1 - 2 x) = 24 x^{2} \times 1 - 24 x^{2} \times 2 x = 24 x^{2} - 48 x^{1 + 2} = 24 x^{2} - 48 x^{3}$
Substituting x = 3 in the result, we get:

24x2−48x3=24(3)2−48(3)3=24×9−48×27=216−1296=−1080 Thus, the product is $(24 x^{2} - 48 x^{3}) and its value fo r x = 3 is (- 1080)$ .

Question 17: Find the product −3y(xy + y²) and find its value for x = 4 and y = 5.

Answer 17: To find the product, we will use distributive law as follows:
$- 3 y (x y + y^{2}) = - 3 y \times x y + (- 3 y) \times y^{2} = - 3 x y^{1 + 1} - 3 y^{1 + 2} = - 3 x y^{2} - 3 y^{3}$ Substituting x = 4 and y = 5 in the result, we get:

−3xy2−3y3=−3(4)(5)2−3(5)3=−3(4)(25)−3(125)=−300−375=−675 Thus, the product is ( $- 3 x y^{2} - 3 y^{3}$ ), and its value for x = 4 and y = 5 is ( $-$ 675).

Question 18: Multiply $- \frac{3}{2} x^{2} y^{3} by (2 x - y)$ and verify the answer for x = 1 and y = 2.

Answer 18: To find the product, we will use distributive law as follows:

−3/2 x2y3×(2x−y)=(−3/2 x2y3×2x)−(−3/2 x2y3×y)=(−3x2+1y3)−(−3/2 x2y3+1)=−3x3y3+3/2 x2y4

Substituting x = 1 and y = 2 in the result, we get:

−3x3y3+3/2 x2y4=−3(1)3(2)3+3/2 (1)2(2)4=−3×1×8+3/2 ×1×16=−24+24=0Thus, the product is $- 3 x^{3} y^{3} + \frac{3}{2} x^{2} y^{4}$ , and its value for x = 1 and y = 2 is 0.

Question 19: Multiply the monomial by the binomial and find the value of each for x = −1, y = 0.25 and z = 0.05:
(i) 15y²(2 − 3x)
(ii) −3x(y² + z²)
(iii) z²(x − y)
(iv) xz(x² + y²)

Answer 19: (i) To find the product, we will use distributive law as follows:
$15 y^{2} (2 - 3 x) = 15 y^{2} \times 2 - 15 y^{2} \times 3 x = 30 y^{2} - 45 x y^{2}$ Substituting x = $-$ 1 and y = 0.25 in the result, we get: 30y2−45xy2=30(0.25)2−45(−1)(0.25)2=30×0.0625−{45×(−1)×0.0625}=30×0.0625−{45×(−1)×0.0625}=1.875−(−2.8125)=1.875+2.8125=4.6875 (ii) To find the product, we will use distributive law as follows:
−3x(y2+z2)=−3x×y2+(−3x)×z2=−3xy2−3xz2 Substituting x = $-$ 1, y = 0.25 and z = 0.05 in the result, we get:

−3xy2−3xz2=−3(−1)(0.25)2−3(−1)(0.05)2=−3(−1)(0.0625)−3(−1)(0.0025)=01875+0.0075=0.195 (iii) To find the product, we will use distributive law as follows:
$z^{2} (x - y) = z^{2} \times x - z^{2} \times y = x z^{2} - y z^{2}$ Substituting x = $-$ 1, y = 0.25 and z = 0.05 in the result, we get:

xz2−yz2=(−1)(0.05)2−(0.25)(0.05)2=(−1)(0.0025)−(0.25)(0.0025)=−0.0025−0.000625=−0.003125 (iv) To find the product, we will use distributive law as follows:

xz(x2+y2)=xz×x2+xz×y2=x3z+xy2z Substituting x = $-$ 1, y = 0.25 and z = 0.05 in the result, we get:
x3z+xy2z=(−1)3(0.05)+(−1)(0.25)2(0.05)=(−1)(0.05)+(−1)(0.0625)(0.05)=−0.05−0.003125=−0.053125

Question 20: Simplify:
(i) 2x²(x³ − x) − 3x(x⁴ + 2x) − 2(x⁴ − 3x²)
(ii) x³y(x² − 2x) + 2xy(x³ − x⁴)
(iii) 3a² + 2(a + 2) − 3a(2a + 1)
(iv) x(x + 4) + 3x(2x² − 1) + 4x² + 4
(v) a(b − c) − b(c − a) − c(a − b)
(vi) a(b − c) + b(c − a) + c(a − b)
(vii) 4ab(a − b) − 6a²(b − b²) − 3b²(2a² − a) + 2ab(b − a)
(viii) x²(x² + 1) − x³(x + 1) − x(x³ − x)
(ix) 2a² + 3a(1 − 2a³) + a(a + 1)
(x) a²(2a − 1) + 3a + a³ − 8

(xii) a²b(a − b²⁾ + ab²(4ab − 2a²) − a³b(1 − 2b)
(xiii) a²b(a³ − a + 1) − ab(a⁴ − 2a² + 2a) − b (a³ − a² − 1)

Answer 20: (i) To simplify, we will use distributive law as follows:
$2 x^{2} (x^{3} - x) - 3 x (x^{4} + 2 x) - 2 (x^{4} - 3 x^{2}) = 2 x^{5} - 2 x^{3} - 3 x^{5} - 6 x^{2} - 2 x^{4} + 6 x^{2} = 2 x^{5} - 3 x^{5} - 2 x^{4} - 2 x^{3} - 6 x^{2} + 6 x^{2} = - x^{5} - 2 x^{4} - 2 x^{3}$ (ii) To simplify, we will use distributive law as follows:
$x^{3} y (x^{2} - 2 x) + 2 x y (x^{3} - x^{4}) = x^{5} y - 2 x^{4} y + 2 x^{4} y - 2 x^{5} y = x^{5} y - 2 x^{5} y - 2 x^{4} y + 2 x^{4} y = - x^{5} y$ (iii) To simplify, we will use distributive law as follows:
3a2+2(a+2)−3a(2a+1)=3a2+2a+4−6a2−3a=3a2−6a2+2a−3a+4=−3a2−a+4 (iv) To simplify, we will use distributive law as follows:

x(x+4)+3x(2x2−1)+4x2+4=x2+4x+6x3−3x+4x2+4=x2+4x2+4x−3x+6x3+4=5x2+x+6x3+4 (v) To simplify, we will use distributive law as follows:

a(b−c)−b(c−a)−c(a−b)=ab−ac−bc+ba−ca+cb=ab+ba−ac−ca−bc+cb=2ab−2ac (vi) To simplify, we will use distributive law as follows:

a(b−c)+b(c−a)+c(a−b)=ab−ac+bc−ba+ca−cb =ab−ba−ac+ca+bc−cb=0 (vii) To simplify, we will use distributive law as follows:

$4 a b (a - b) - 6 a^{2} (b - b^{2}) - 3 b^{2} (2 a^{2} - a) + 2 a b (b - a) = 4 a^{2} b - 4 a b^{2} - 6 a^{2} b + 6 a^{2} b^{2} - 6 b^{2} a^{2} + 3 b^{2} a + 2 a b^{2} - 2 a^{2} b = 4 a^{2} b - 6 a^{2} b - 2 a^{2} b - 4 a b^{2} + 3 b^{2} a + 2 a b^{2} + 6 a^{2} b^{2} - 6 b^{2} a^{2} = - 4 a^{2} b + a b^{2}$ (viii) To simplify, we will use distributive law as follows:

x2(x2+1)−x3(x+1)−x(x3−x)=x4+x2−x4−x3−x4+x2=x4−x4−x4−x3+x2+x2=−x4−x3+2x2 (ix) To simplify, we will use distributive law as follows:

2a2+3a(1−2a3)+a(a+1)=2a2+3a−6a4+a2+a=2a2+a2+3a+a−6a4=3a2+4a−6a4 (x) To simplify, we will use distributive law as follows:

a2(2a−1)+3a+a3−8=2a3−a2+3a+a3−8=2a3+a3−a2+3a−8=3a3−a2+3a−8 (xi) To simplify, we will use distributive law as follows:

RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) | RD Sharma Solutions for Class 8 Mathematics

(xii) To simplify, we will use distributive law as follows:

a2b(a−b2)+ab2(4ab−2a2)−a3b(1−2b)=a3b−a2b3+4a2b3−2a3b2−a3b+2a3b2=a3b−a3b−a2b3+4a2b3−2a3b2+2a3b2=3a2b3 (xiii) To simplify, we will use distributive law as follows:

a2b(a3−a+1)−ab(a4−2a2+2a)−b(a3−a2−1)=a5b−a3b+a2b−a5b+2a3b−2a2b−a3b+a2b+b=a5b−a5b−a3b+2a3b−a3b+a2b−2a2b+a2b+b=b

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RD Sharma Solutions for Class 8 Math Chapter 6 - Algebraic Expressions and Identities (Part-3) | RD Sharma Solutions for Class 8 Mathematics PDF Download

Question 1: Find the following product:2a3(3a + 5b)

Answer 1: To find the product, we will use distributive law as follows:2a3(3a+5b)=2a3×3a+2a3×5b=(2×3)(a3×a)+(2×5)a3b=(2×3)a3+1+(2×5)a3b=6a4+10a3b2a33a+5b=2a3×3a+2a3×5b=2×3a3×a+2×5a3b=2×3a3+1+2×5a3b=6a4+10a3bThus, the answer is 6a4+10a3b6a4+10a3b.

Question 2: Find the following product:−11a(3a + 2b)

Question 3: Find the following product:−5a(7a − 2b)

Question 4: Find the following product:−11y2(3y + 7)

Answer 4: To find the product, we will use distributive law as follows:−11y2(3y+7)=(−11y2)×3y+(−11y2)×7=(−11×3)(y2×y)+(−11×7)×(y2)=(−33)(y2+1)+(−77)×(y2)=−33y3−77y2

Thus, the answer is −33y3−77y2.

Question 5: Find the following product:6x/5 (x3+y3)6x5(x3+y3)

Answer 5: To find the product, we will use distributive law as follows:

Question 6: xy(x3 − y3)

Answer 6: To find the product, we will use the distributive law in the following way:xy(x3−y3)=xy×x3−xy×y3=(x×x3)×y−x×(y×y3)=x1+3y−xy1+3=x4y−xy4xyx3-y3=xy×x3-xy×y3=x×x3×y-x×y×y3=x1+3y-xy1+3=x4y-xy4Thus, the answer is x4y−xy4x4y-xy4.

Question 7: Find the following product:0.1y(0.1x5 + 0.1y)

Question 8: Find the following product:

Answer 8: To find the product, we will use distributive law as follows:

Question 9: Find the following product:

Answer 9: To find the product, we will use the distributive law in the following way:

Question 10: Find the following product:

Answer 10: To find the product, we will use distributive law as follows:

Question 11: Find the following product:1.5x(10x2y − 100xy2)

Answer 11: To find the product, we will use distributive law as follows:1.5x(10x2y−100xy2)=(1.5x×10x2y)−(1.5x×100xy2)=(15x1+2y)−(150x1+1y2)=15x3y−150x2y2Thus, the answer is 15x3y−150x2y215x3y-150x2y2.

Question 12: Find the following product:4.1xy(1.1x − y)

Answer 12: To find the product, we will use distributive law as follows:4.1xy(1.1x−y)=(4.1xy×1.1x)−(4.1xy×y)={(4.1×1.1)×xy×x}−(4.1xy×y)=(4.51x1+1y)−(4.1xy1+1)=4.51x2y−4.1xy2Thus, the answer is 4.51x2y−4.1xy24.51x2y-4.1xy2.

Question 13: Find the following product:

Answer 13: To find the product, we will use distributive law as follows:

Question 14: Find the following product:

Answer 14: To find the product, we will use distributive law as follows:

Question 15: Find the following product:

Answer 15: To find the product, we will use distributive law as follows:

Question 16: Find the product 24x2 (1 − 2x) and evaluate its value for x = 3.

Answer 16: To find the product, we will use distributive law as follows:24x2(1−2x)=24x2×1−24x2×2x=24x2−48x1+2=24x2−48x324x21-2x=24x2×1-24x2×2x=24x2-48x1+2=24x2-48x3Substituting x = 3 in the result, we get:

Question 17: Find the product −3y(xy + y2) and find its value for x = 4 and y = 5.

Answer 17: To find the product, we will use distributive law as follows:−3y(xy+y2)=−3y×xy+(−3y)×y2=−3xy1+1−3y1+2=−3xy2−3y3Substituting x = 4 and y = 5 in the result, we get:

Question 18: Multiply −3/2 x2y3 by (2x − y)-32x2y3 by (2x - y) and verify the answer for x = 1 and y = 2.

Answer 18: To find the product, we will use distributive law as follows:

Question 19: Multiply the monomial by the binomial and find the value of each for x = −1, y = 0.25 and z = 0.05:(i) 15y2(2 − 3x)(ii) −3x(y2 + z2)(iii) z2(x − y)(iv) xz(x2 + y2)

−3xy2−3xz2=−3(−1)(0.25)2−3(−1)(0.05)2=−3(−1)(0.0625)−3(−1)(0.0025)=01875+0.0075=0.195 (iii) To find the product, we will use distributive law as follows:z2(x−y)=z2×x−z2×y=xz2−yz2z2x-y=z2×x-z2×y=xz2-yzSubstituting x = −-1, y = 0.25 and z = 0.05 in the result, we get:

xz2−yz2=(−1)(0.05)2−(0.25)(0.05)2=(−1)(0.0025)−(0.25)(0.0025)=−0.0025−0.000625=−0.003125 (iv) To find the product, we will use distributive law as follows:

xz(x2+y2)=xz×x2+xz×y2=x3z+xy2z Substituting x = −-1, y = 0.25 and z = 0.05 in the result, we get:x3z+xy2z=(−1)3(0.05)+(−1)(0.25)2(0.05)=(−1)(0.05)+(−1)(0.0625)(0.05)=−0.05−0.003125=−0.053125

(xii) a2b(a − b2) + ab2(4ab − 2a2) − a3b(1 − 2b)(xiii) a2b(a3 − a + 1) − ab(a4 − 2a2 + 2a) − b (a3 − a2 − 1)

x(x+4)+3x(2x2−1)+4x2+4=x2+4x+6x3−3x+4x2+4=x2+4x2+4x−3x+6x3+4=5x2+x+6x3+4 (v) To simplify, we will use distributive law as follows:

a(b−c)−b(c−a)−c(a−b)=ab−ac−bc+ba−ca+cb=ab+ba−ac−ca−bc+cb=2ab−2ac (vi) To simplify, we will use distributive law as follows:

a(b−c)+b(c−a)+c(a−b)=ab−ac+bc−ba+ca−cb =ab−ba−ac+ca+bc−cb=0 (vii) To simplify, we will use distributive law as follows:

4ab(a−b)−6a2(b−b2)−3b2(2a2−a)+2ab(b−a)=4a2b−4ab2−6a2b+6a2b2−6b2a2+3b2a+2ab2−2a2b=4a2b−6a2b−2a2b−4ab2+3b2a+2ab2+6a2b2−6b2a2=−4a2b+ab2(viii) To simplify, we will use distributive law as follows:

x2(x2+1)−x3(x+1)−x(x3−x)=x4+x2−x4−x3−x4+x2=x4−x4−x4−x3+x2+x2=−x4−x3+2x2 (ix) To simplify, we will use distributive law as follows:

2a2+3a(1−2a3)+a(a+1)=2a2+3a−6a4+a2+a=2a2+a2+3a+a−6a4=3a2+4a−6a4 (x) To simplify, we will use distributive law as follows:

a2(2a−1)+3a+a3−8=2a3−a2+3a+a3−8=2a3+a3−a2+3a−8=3a3−a2+3a−8 (xi) To simplify, we will use distributive law as follows:

RD Sharma Solutions for Class 8 Mathematics

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Question 1: Find the following product:
2a³(3a + 5b)

Question 2: Find the following product:
−11a(3a + 2b)

Question 3: Find the following product:
−5a(7a − 2b)

Question 4: Find the following product:
−11y²(3y + 7)

Thus, the answer is −33y³−77y².

Question 5: Find the following product:
$\frac{6 x}{5} (x^{3} + y^{3})$

Question 6: xy(x³ − y³)

Answer 6: To find the product, we will use the distributive law in the following way:
$x y (x^{3} - y^{3}) = x y \times x^{3} - x y \times y^{3} = (x \times x^{3}) \times y - x \times (y \times y^{3}) = x^{1 + 3} y - x y^{1 + 3} = x^{4} y - x y^{4}$
Thus, the answer is $x^{4} y - x y^{4}$ .

Question 7: Find the following product:
0.1y(0.1x⁵ + 0.1y)

Question 11: Find the following product:
1.5x(10x²y − 100xy²)

Answer 11: To find the product, we will use distributive law as follows:
$1.5 x (10 x^{2} y - 100 x y^{2}) = (1.5 x \times 10 x^{2} y) - (1.5 x \times 100 x y^{2}) = (15 x^{1 + 2} y) - (150 x^{1 + 1} y^{2}) = 15 x^{3} y - 150 x^{2} y^{2}$ Thus, the answer is $15 x^{3} y - 150 x^{2} y^{2}$ .

Question 12: Find the following product:
4.1xy(1.1x − y)

Question 16: Find the product 24x² (1 − 2x) and evaluate its value for x = 3.

Answer 16: To find the product, we will use distributive law as follows:
$24 x^{2} (1 - 2 x) = 24 x^{2} \times 1 - 24 x^{2} \times 2 x = 24 x^{2} - 48 x^{1 + 2} = 24 x^{2} - 48 x^{3}$
Substituting x = 3 in the result, we get:

Question 17: Find the product −3y(xy + y²) and find its value for x = 4 and y = 5.

Answer 17: To find the product, we will use distributive law as follows:
$- 3 y (x y + y^{2}) = - 3 y \times x y + (- 3 y) \times y^{2} = - 3 x y^{1 + 1} - 3 y^{1 + 2} = - 3 x y^{2} - 3 y^{3}$ Substituting x = 4 and y = 5 in the result, we get:

Question 18: Multiply $- \frac{3}{2} x^{2} y^{3} by (2 x - y)$ and verify the answer for x = 1 and y = 2.

Question 19: Multiply the monomial by the binomial and find the value of each for x = −1, y = 0.25 and z = 0.05:
(i) 15y²(2 − 3x)
(ii) −3x(y² + z²)
(iii) z²(x − y)
(iv) xz(x² + y²)

xz(x2+y2)=xz×x2+xz×y2=x3z+xy2z Substituting x = $-$ 1, y = 0.25 and z = 0.05 in the result, we get:
x3z+xy2z=(−1)3(0.05)+(−1)(0.25)2(0.05)=(−1)(0.05)+(−1)(0.0625)(0.05)=−0.05−0.003125=−0.053125

(xii) a²b(a − b²⁾ + ab²(4ab − 2a²) − a³b(1 − 2b)
(xiii) a²b(a³ − a + 1) − ab(a⁴ − 2a² + 2a) − b (a³ − a² − 1)