# 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

1. 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

1. 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

1. 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

1. 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.

1. 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

1. 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 – 3x2 𝟏/𝐱 + 3x(𝟏/𝐱)3 – (𝟏/𝐱)3 = x3 – 3x + 3x/ x21/x3

= x3 – 3x + 3/x1/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

1. If x + 𝟏/𝐱 = 3, prove that x3 + 𝟏/𝐱𝟑 = 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

1. 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

1. 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

1. If a2 + 𝟏/𝐚𝟐 = 20 and a3 + 𝟏/𝐚𝟑 = 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

## 3 thoughts on “Multiplication of Polynomials – Exercise 3.1.3-Class 9”

1. mark schwartz says:

is the poly mult a Vedic method? if so, what’s the source? I have Vedic Math,1992, and don’t recall seeing that method in that book …

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1. breathmath says:

No, Sir.

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