Exercise 6.4
Find the following products:
1. 2a3 (3a+5b)
Solution:
To find the product, we will use distributive law as follows:
2a3 (3a + 5b)
= 2a3 × 3a + 2a3 × 5b
= (2 x 3)(a3 × a) + (2 × 5)a3b
= (2×3)a3+1+ (2×5)a3b
= 6a4 +10a3b
Thus, the answer is 6a4 +10a3b.
2. -11a (3a + 2b)
Solution:
To find the product, we will use distributive law as follows:
-11a(3a + 2b)
= (-11a) × 3a + (-11a) × 2b
= (-11 × 3) × (a × a) + (-11 × 2) × (a × b)
= (-33) × (a1+1) + (-22) × (a × b)
= -33a2 – 22ab
Thus, the answer is -33a2 – 22ab.
3. -5a (7a – 2b)
Solution:
To find the product, we will use distributive law as follows:
-5a(7a – 2b)
= (-5a) × 7a + (-5a) × (-2b)
= (-5 × 7) × (a × a) + (-5 × (-2)) × (a × b)
= (-35) × (a1+1) + (10) × (a × b)
= -35a2 + 10ab
Thus, the answer is -35a2 + 10ab.
4. -11y2 (3y + 7)
Solution:
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.
6. xy (x3 – y3)
Solution:
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 – xy4
Thus, the answer is x4y – xy4
7. 0.1y (0.1×5 + 0.1y)
Solution:
To find the product, we will use distributive law as follows:
0.1y (0.1×5 + 0.1y)
= (0.ly) (0.1x5) + (0.ly) (0.ly)
= (0.1 × 0.1) (y × x5) + (0.1 × 0.1) (y × y)
= (0.1 × 0.1) (x5 × y) + (0.1 × 0.1) ( y1+1)
= 0.01x5y + 0.01y2
Thus, the answer is 0.01x5y + 0.01y2
11. 1.5x (10x2y–100xy2)
Solution:
To find the product, we will use distributive law as follows:
=(1.5x×10x2y)–(1.5x×100xy2)
=(15x1+2y)–(150x1+1y2)
=15x3y–150x2y2
Thus, the answer is 15x3y–150x2y2
12. 4.1 xy (1.1x – y)
Solution:
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)
= (4.51x 1+1 y)−(4.1 xy1 +1)
= 4.51x2y–4.1xy2
Thus, the answer is 4.51x2y–4.1xy2
16. Find the product 24x2(1–2x)and evaluate its value for x = 3.
Solution:
To find the product, we will use distributive law as follows:
24x2(1–2x)
= 24x2 × 1–24x2 × 2x
= 24x2 – 48x1+2
= 24x2 – 48x3
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 24x2–48x3 and its value for x = 3 is -1080.
17. Find the product −3y (xy+y2)and find its value for x = 4 and y = 5.
Solution:
To find the product, we will use distributive law as follows:
−3y (xy+y2)
= −3y × xy + (−3y) × y2
=−3xy1+1 – 3y1+2
=−3xy2–3y3
Substituting x = 4 and y = 5 in the result, we get
−3xy2–3y
= −3 (4) (5)2–3(5)3
= −3 (4) (25)–3(125)
=−300 – 375
= − 675
Thus, the product is −3xy2–3y3, and its value for x = 4 and y = 5 is -675.
19. Multiply the monomial by the binomial and find the value of each for x = -1, y = 0.25 and z = 0.05:
Solution:
(i) 15y2 (2–3x)
To find the product, we will use distributive law as follows:
15y2 (2–3x)
= 15y2 × 2 – 15y2 × 3x
= 30y2–45xy2
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}
=1.875 – (−2.8125)
=1.875 + 2.8125
=4.6875
20. Simplify:
(i) 2×2(x3–x)–3x(x4+2x)–2(x4–3x2)
Solution
To simplify, we will use distributive law as follows:
2x2(x3–x)–3x(x4+2x)–2(x4–3x2)
=2x5–2x3–3x5–6x2–2x4+6x2
=2x5–3x5–2x4–2x3–6x2+6x2
=−x5–2x4–2x3
(ii) x3y(x2–2x)+2xy(x3–x4)
Solution:
To simplify, we will use distributive law as follows:
x3y(x2–2x)+2xy(x3–x4)
=x5y–2x4y+2x4y–2x5y
=x5y–2x5y–2x4y+2x4y
=−x5y
(iii) 3a2+2(a+2)–3a(2a+1)
Solution:
To simplify, we will use distributive law as follows:
3a2+2(a+2) –3a (2a+1)
=3a+2a+4–6a2–3a
=3a2–6a2–3a+4
=–3a2–a+4
(iv) x(x+4)+3x(2x2–1)+4x2+4
Solution:
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) a (b – c) – b(c – a) – c(a – b)
Solution:
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
(vi) a(b – c) + b(c – a) + c(a – b)
Solution:
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) 4ab(a–b)–6a2(b–b2)–3b2(2a2–a)+2ab(b–a)
Solution:
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) x2(x2+1)–x3(x+1)–x(x3–x)
Solution:
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) 2a2+3a(1–2a3)+a(a+1)
Solution:
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
(x) a2(2a–1)+3a+a3–8
Solution:
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) 32×2(x2–1)+14×2(x2+x)–34x(x3–1)
Solution:
To simplify, we will use distributive law as follows:
32x2(x2–1)+14x2(x2+x)–34x(x3–1)
=32x4–32x2+14x4+14x3–34x4+34x
=32x4+14x4–34x4+14x3–32x2+34x
=x4+14x3–32x2+34x
(xii) a2b(a–b2)+ab2(4ab–2a2)–a3b(1–2b)
Solution:
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) a2b(a3–a+1)–ab(a4–2a2+2a)–b(a3–a2–1)
Solution:
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–2a2b+a2b+b
=b