Multiplication of Polynomials – Exercise 3.1.3-Class 9

1. Find the following products:

(i)(x + 4) (x + 5) (x + 2)

Using (x + a) (x + b) (x + c) = x³ + x² (a + b + c) + x (ab + bc + ca) + abc

we get

a = 4, b = 5 and c =2

(x + 4) (x + 5) (x + 2) = x³ + x²(4 + 5 + 2) + x (4.5 + 5.2 + 2.4) + 4.5.2

= x³ + 11x² + x (20 + 10 + 8) +40

= x³ + 11x² + 38x +40

 

(ii) (y + 3) (y + 2) ( y – 1)

Using (x + a) (x + b) (x + c) = x³ + x² (a + b + c) + x (ab + bc + ca) + abc we get

x = y, a = 3, b = 2 and c = -1

(y + 3) (y + 2) (y – 1) = y³ + y² (3 + 2 – 1) + y (3.2 + 2(-1) + (-1)3 + 3.2(-1)

= y³ + 4y² + y (6 – 2 – 3) – 6

= y³ + 4y² + y – 6

 

(iii) (a + 2) (a – 3) (a + 4)

Using (x + a) (x + b) (x + c) = x³ + x² (a + b + c) + x (ab + bc + ca) + abc

we get

x = a, a = 2, b = –3 and c = 4

(a + 2) (a – 3) (a – 4) = a³ + a² (2 – 3 + 4) + a [2(–3) + (–3)4 + 4.2) + 2 (–3) 4

= a³ + 3a² + a (–6 – 12 + 8) – 24

= a³ + 3a² – 10a – 24

 

(iv) (m – 1) (m – 2) (m – 3)

Using (x + a) (x + b) (x + c) = x³ + x² (a + b + c) + x (ab + bc + ca) + abc

we get

x = m, a = –1, b = –2 and c = –3

(m – 1) (m – 2) (m – 3) = m³ + m² (–1 – 2 – 3) + m [(–1) (–2) + (–2) (–3) +

(– 3)(–1)] + (–1) (–2) (–3)

= m³ + m² (–6) + m [2 + 6 + 3] – 6

= m³ – 6m² + 11m – 6

 

(v) ( √𝟐 + √𝟑) (√𝟐+ √𝟓) (√𝟐+ √𝟕 )

Using (x + a) (x + b) (x + c) = x³ + x² (a + b + c) + x (ab + bc + ca) + abc

we get

x = 2, a = 3, b = 5 and c = 7

(√2 + √3) (√2+ √5) (√2+ √7 ) = (√2)³ + (√2)² [√3 + √5+ √7]+ √2 [√3. √5 + √5. √7 + √7. √3] + √3. √5. √7

= 2√2 + 2(√3 + √5 + √7) + √2 (15+ √35 + √21) + √105

 

(vi) 105 x 101 x 102

We can write this as

(100 + 5) (100 + 1) (100 + 2)

Using (x + a) (x + b) (x + c) = x³ + x² (a + b + c) + x (ab + bc + ca) + abc

we get

x = 100, a = 5, b = 1 and c = 2

(100 + 5) (100 + 1) (100 + 2) = 100³ + 100² (5 + 1 + 2) + 100 (5.1 + 1.2

+ 2.5) + 5.1.2

= 1000000 + 10000 (8) + 100(5 + 2 +10) + 10

= 1000000 + 80000 + 1700 +10

= 1081710

 

(vii) 95 x 98 x 103

We can write this as

(100 – 5) (100 – 2) (100 + 3)

Using (x + a) (x + b) (x + c) = x³ + x² (a + b + c) + x (ab + bc + ca) + abc

we get

x = 100, a = -5, b = -2 and c = 3

(100 – 5) (100 – 2) (100 + 3) = 100³ + 100² (–5 – 2 +3) + 100(–5) (–2) + (–2) 3

+ 3 (-5) + (-5) (-2) 3

= 1000000 + 10000(–4) +100 (10 – 6 –16) + 30

= 1000000 – 40000 – 1100 +30

= 958930

 

(viii) 1.01 x 1.02 x 1.03

We can write this as

(1 + 0.01) (1 + 0.02) (1 + 0.03)

Using (x + a) (x + b) (x + c) = x³ + x² (a + b + c) + x (ab + bc + ca) + abc

we get

x = 1, a = 0.01, b = 0.02 and c = 0.03

(1 + 0.01) (1 + 0.02) (1 + 0.03) = 1³ + 1² (0.01 + 0.02 + 0.03) + 1 [(0.01) (0.02) +

(0.02) (0.03) + (0.03) (0.01)] + (0.01) (0.02) (0.03)

= 1 + 0.06 + (0.0002 + 0.0006 + 0.0003) + 0.000006

= 1.061106

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2. Find the coefficients of x2 and x in the following:

(i) (x + 4) (x + 1) (x + 2)

= x³ + x² (4 + 1 + 2) + x (4.1 + 1.2 + 2.4) + 4.1.2

= x³ + 7x² + 14x + 8

Coefficient of x² is 7

Coefficient of x is 14

 

(ii) (x – 5) (x – 6) (x – 1)

= x³ + x² (–5 – 6 –1) + x [(–5) (–6) + (–6) (–1) – 1(–5)] + (–5) (–6) (–1)

= x³– 12x² + x (30 + 6 + 5) – 30

= x³– 12x² + 41x – 30

Coefficient of x² is –12 and x is 41

 

(iii) (2x + 1) (2x – 2) (2x – 5)

= (2x)³ + (2x)² [1–2–5] + 2x [(1)(–2) + (–2)(–5) + (–5)(1)] + 1 (–2) (–5)

= 8x³ + 4x² (–6) + 2x [–2 +10–5] + 10

= 8x³ – 24x² + 6x + 10

Coefficient of x² is –24 and x is 6

 

(iv) ( ^𝐱/_𝟐 + 1) ( ^𝐱/_𝟐 + 2) ( ^𝐱/_𝟐 + 3)

= (^𝐱/ )3 + (^𝐱/_𝟐)2 [1 + 2 + 3] + ^𝐱/_𝟐 [1.2 + 2.3 + 3.1] + 1.2.3

= x³/₈ + (x²/₄ )(6) + ^𝐱/_𝟐 (2 + 6 + 3) + 6

= x³/₈ + ³/₂ x² + ¹¹/₂ x + 6

Coefficient of x² is 𝟑𝟐 and x is 𝟏𝟏𝟐

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3. The length, breadth and height of a cuboids are (x +3), (x – 2) and (x -1) respectively. Find its volume.

Solution:

Volume of a cuboids = length x breadth x height

V = (x +3) (x – 2) (x – 1)

Using the identity (x + a) (x + b) (x + c) = x³ + x² (a + b + c) + x (ab + bc + ca) + abc we get

V = x³ + x² (3 – 2 – 1) + x [3(–2) + (–2) (–1) + (–1)3] + 3(–2) (–1)

= x³ – 0x² + X (–6 + 2 – 3) + 6

= x³ – 7x² + 6

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4. The length, breadth and height of a metal box are cuboid are (x +5), (x – 2) and (x – 1) respectively. What is its volume?

Solution:

Volume of the metal box = length x breadth x height

V = (x +5) (x – 2) (x – 1)

Using the identity (x + a) (x + b) (x + c) = x³ + x² (a + b + c) + x (ab + bc + ca) + abc we get

V = x³ + x² + (–5 – 2 –1) + x [5(–2) + (–2) (–1) + (–1)5] + 5 (–2) (–1)

= x³ + 2x² + x [–10 + 2 – 5] + 10

= x³ + 2x² – 13x + 10

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5. Prove that

(a + b) (b + c) (c + a) = (a + b + c) (ab + bc + ca) – abc

[Hint: write a + b = a + b + c – c, b + c = a + b + c – a, c + a = a + b + c – d]

Solution:

x³ + x² (a + b + c) + x (ab + bc + ca) + abc we get

1. H.S. = (a + b + c)³ + (a + b + c)² (–c – a – b) + (a + b + c) [(–c) (–a) +

(–a) (–b) + (–b) (–c)] – (–c) (–a) (–b)

= (a + b + c)³-(a + b + c)²[(a + b + c)] +(a + b + c) (ac + ab + bc) – abc

= (a + b + c)³– (a + b + c)³ + (a + b + c) (ac + ab + bc) – abc

= (a + b + c) (ac + ab + bc) – abc

= R. H. S.

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6. Find the cubes of the following:

(i) (2x +y)³

Solution:

Using (a + b)³ = a³ + 3a²b + 3ab² + b³ we get

a = 2x, b = y

(2x +y)³ = (2x)³ + 3(2x)² y + 3 (2x) y² +y³

= 8x³ + 12 x²y + 6 xy² +y³

 

(ii) (2x + 3y)³

Solution:

Using (a + b)³ = a³ + 3a²b + 3ab² + b³ we get

a = 2x, b = 3y

(2x + 3y)³ = (2x)³ + 3(2x)² (3y) + 3 (2x) (3y)² + (3y)³

= 8x³ + 36 x²y + 54 xy² + 27y³

 

(viii) 101³

Solution:

We write 101 as (100 + 1)³

Using identity (a + b)³ = a³ + 3a²b + 3ab² + b³ we get

a = 100, b = 1

(100 + 1)³ = 100³ + 3. 100² + 3. 100.1² + 1³

= 1000000 + 30000 + 300 + 1

= 1030301

 

(viii) 2.1³

Solution:

We write 2.1 (2 + 0.1)³

Using identity (a + b)³ = a³ + 3a²b + 3ab² + b³ we get

a = 2, b = 0.1

(2 + 0.1)³ = 2³ + 3 x 2²(0.1) + 3 x 2 x (0.1)² + (0.1)³

= 8 + 1.2 + 0.06 + 0.001

= 9.261

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7. Find the cubes of the following:

(i) (2a – 3b)³

Solution:

Using (a – b)³ = a³ – 3a²b + 3ab² – b³ we get

a = 2a, b = 3b

(2a – 3b)³ = (2a)³ – 3 (2a)²(3b) + 3 (2a)(3b)² – (3b)³

= 8a³ – 36a²b + 54ab² -27b³

 

(ii) ( x – ^𝟏/_𝐱 )³

Solution:

Using (a – b)³ = a³ – 3a²b + 3ab² – b³ we get

a = x, b = ^𝟏/_𝐱

(x – ^𝟏/ )³ = x³ – 3x^{2 𝟏}/_𝐱 + 3x(^𝟏/_𝐱)³ – (^𝟏/_𝐱)³ = x³ – 3x + ^3x/ x² – ¹/x³

= x³ – 3x + ³/_x – ¹/x³

 

(iii) (√3 x – 2)²

Solution:

Using (a – b)³ = a³ – 3a²b + 3ab² – b³ we get

a = √3 x, b = 2

(√3 x – 2)² = (√3x)³ – 3 (√3x)² .2 + 3. √3x x 2² – 2³

= 3√3 x³ – 6. 3 x² + 12√3 x – 8

=3√3 x³ – 18x² + 12√3 x – 8

 

(iv) (2x – √5)³

Solution:

Using (a – b)³ = a³ – 3a²b + 3ab² – b³ we get

a = 2x, b = √5

(2x – √5)³ = (2x)³ – 3(2x)² √5 + 3. 2x. √5)² – (√5)³

= 8x³ – 12√5x² + 30x – 5√5

 

(v) 49³

Solution:

We can write 49 = 50 – 1

Using (a – b)³ = a³ – 3a²b + 3ab² – b³ we get

a = 50, b = 1

(50 – 1)³ = 50³ – 3.50².1 + 3.50.1² – 1³

= 125000 – 3 x 2500 + 150 – 1

= 125000 – 7500 +149

= 117649

 

(vi) 18³

Solution:

Let us write 18 = 20 – 2

Using (a – b)³ = a³ – 3a²b + 3ab² – b³ we get

a = 20, b = 2

(20 – 2)³ = 20³ – 3.20².2 + 3.20.2² – 2³

= 8000 – 6×400 + 60×4 – 8

= 8000 – 2400 +240 – 8

= 5832

 

(vii) 95³

Solution:

We write 95 = 100 – 5

Using (a – b)³ = a³ – 3a²b + 3ab² – b³ we get

a = 100, b = 5

(100 – 5)³ = 100³ – 3.100².5 + 3.100.5² – 5³

= 1000000 – 150000 + 7500 – 125

= 857375

 

(viii) 108³

Solution:

We write 108³ = (110 – 2)

Using (a – b)³ = a³ – 3a²b + 3ab² – b³ we get

a = 110, b = -2

(110 – 2)³ = 110³ – 3. (110)².2 + 3.110×2² – 2³

= 1331000 – 72600 + 1320 -8

= 1259712

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8. If x + ^𝟏/_𝐱 = 3, prove that x³ + 𝟏/𝐱^𝟑 = 18.

Solution:

Given x + ¹/_x = 3

Cubing both sides we get

(x + ¹/_x )³ = 3³

Using (a + b)³ = a³ + 3a²b + 3ab² + b³ we get

a = x b = 1/x

(x + 1/x )³ = (x)³ + ( 1/x )³ + 3x. ¹/_x (x + ¹/_x )

27 = x³ + ¹/x³ + 3 (3)

x³ + 1/x³ = 27 – 9

= x³ + ¹/x³ = 18

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9. If p + q = 5 and pq = 6, find p³ + q³

Solution:

(p + q)³ = p³ + 3pq (p + q) + q³

5³ = p³ + 3.6 (5) + q³

125 = p³ + 90 + q³

p³ + q³ = 125 – 90

p³ + q³ = 35

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10. If a – b = 3 and ab = 10, find a³ – b³

Solution:

Given a – b = 3 and ab = 10

(a – b)³ = a³ – b³ – 3ab (a – b)

3³ = a³ – b³ – 3.10 (3)

27 = a³ – b³ – 90

a³ – b³ = 27 + 90

a³ – b³ = 117

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11. If a² + ^𝟏/𝐚^𝟐 = 20 and a³ + ^𝟏/𝐚^𝟑 = 30, find a + 𝟏/𝐚

Solution:

a³ + 1/a³ = (a + ¹/_a) (a² + ¹/a² – a x ¹/_a )

30 = (a + ¹/_a ) = (20 – 1)

30 = (a +¹/_a ) x 19

³⁰/₁₉ = a + ¹/_a

a + ¹/_a = ^𝟑𝟎/_𝟏𝟗

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