Multiplication of Polynomials – Exercise 3.1.2-Class 9

  1. Evaluate the following products:

(i) ax2 ( bx + c)

= ax2 (bx) + ax2 (c)

= abx3 + acx2

 

(ii) ab (a+b)

= ab (a) + ab (b)

= a2b + ab2

 

(iii) a2b2 (ab2+a2b)

= a2b2 (ab2) + a2b2 (a2b)

= a3b4 + a4b3

 

(iv) b4(b6 + b8)

= b4 (b6) + b4 (b8)

= b4 (b6) + b4 (b8)

= b10 + b12


  1. Evaluate the following products:

(i) (x+3) (x+2)

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

= x2 + 3x + 2x + 6

= x2 + 5x + 6

 

(ii) (x+5) (x–2)

= (x+5) x + (x+5) (–2)

= x2 +5x – 2x –10

= x2 + 3x –10

 

(iii) ( y – 4 ) ( y + 6 )

= (y – 4) y +(y – 4)6

= y2 – 4y + 6y – 24

= y2 + 2y – 24

 

(iv) (a–5) (a–6)

= (a–5) a + (a–6) (–6)

= a2 – 5a – 6a + 30

= a2 – 11a +30

 

(v) (2x+1) (2x–3)

= (2x+1)2x + (2x+1) (–3)

= 4x2 + 2x – 6x – 3x

= 4x2 – 4x –3

 

(vi) ( a + b ) ( c + d )

= (a + b) c + (a+b) d

= ac + bc + ad + bd

 

(vii) ( 2x – 3y ) ( x – y )

= (2x – 3y) x + (2x – 3y) (–y)

= 2x2 – 3xy – 2xy +3y2

= 2x2 – 5xy + 3y2

 

(viii) ( √𝟕𝐱 + √𝟓 ) (√𝟓𝐱 + √𝟕 )

= (√𝟕𝐱 + √𝟓 ) (√𝟓𝐱) + (√𝟕𝐱 + √𝟓 ) (√𝟕)

= √𝟑𝟓 x2 + 5x + 7x + √𝟑𝟓

= √𝟑𝟓x2 + 12x + √𝟑𝟓

 

(xi) (2a+3b) (2a–3b)

= (2a+3b) 2a + (2a+3b) (–3b)

= 4a2+ 6ab – 6ab – 9b2

= 4a2 – 9b2

 

(xii) (6xy–5) (6xy+5)

= (6xy – 5) (6xy) + (6xy – 5) 5

= 36x2y2 – 30xy + 30xy – 25

= 36x2y2 – 25

 

(xiii)(2/x+3) (2/x –7)

= (2/x+3)2/x +(2/x+3)(-7)

= 4/x2 + 6/x14/x – 21

= 4/x28/x –21


  1. Expand the following using appropriate identity:

(i) (a +5)2

Using (a + b)2 = a2 +2ab +b2 we get

a = a b = 5

(a +5)2 = a2 +2.a.5 +b2

= a2 +10a +25

 

(ii) (2a +3)2

Using (a + b)2 = a2 +2ab +b2 we get

a = 2a; b = 3

(2a +3)2 = (2a)2 +2.2a.3 +b2

= 4a2 + 12a + 9

 

(iii) ( x + 𝟏/𝐱 )2

Using (a + b)2 = a2 +2ab +b2 we get

a = 2a; b = 1/x

(X + 𝟏/ )2 = x2 + 2.x. 1x + (𝟏/𝐱 )2

= x2 + 2 + (𝟏/x )2

 

(iv) ( √(12a) + √(6b) )2

Using (a + b)2 = a2 +2ab +b2 we get

a = √(12a) and b = √(6b)

(√(12a) + √ (6b))2= (√12a)2 + 2.√12 a + √(6b) + (√(6b))2

= 12a2 +2√72 ab +6b2

= 12a2 +2 √ (36 ×2) ab +6b2

= 12a2 + 12 √𝟐ab +6b2

 

(v) (𝛑 + 𝟐𝟐/7 )2

Using (a + b)2 = a2 +2ab +b2 we get

a = 𝛑 and b = 22/7

(𝛑 + 22/7 )2 = 𝛑2 + 2. 𝛑 22/7 + (22/7)2

= 𝛑2 + 44π/7 + ( 22/7)2

= 𝛑2 + 𝟒𝟒𝛑/7 + 𝟒𝟖𝟒/49

 

(vi) (y – 3)2

Using (a – b)2 = a2 – 2ab + b2

a = y and b = –3

(y – 3)2 = y2 – 2. y.3 + 32

= y2 – 6y + 9

 

(vii) (3a – 2b)2

Using (a – b)2 = a2 – 2ab + b2

a = 3a and b = –2b

(3a – 2b)2 = (3a)2 – 2.3a.2b + (2b)2

= 9a2 – 12ab + 4b2

 

(viii) ( y – 𝟏/y )2

Using (a – b)2 = a2 – 2ab + b2

a = y and b = 𝟏/y

(y – 𝟏/y)2 = y2 – 2.y. 𝟏/y + (𝟏/y )2

= y2 – 2 + 𝟏/

 

(ix) ( √(10x) – √(5y))2

Using (a – b)2 = a2 – 2ab + b2

a = √(10x) and b = √(5y)

(√(10x) – √(5y))2

= (√(10x))2 – 2. √(10x).√(5y)+ √(5y))2

= 10x2 – 2√50 xy + 5y2

= 10x2 – 2.5√2 xy + 5y2

= 10x2 – 10√2 xy + 5y2

 

(x) (𝛑 – 𝟐𝟐/7 )2

Using (a + b)2 = a2 +2ab +b2 we get

a = 𝛑 and b = 𝟐𝟐/𝟕

(𝛑 – 𝟐𝟐/ )2 = 𝛑2 – 2. 𝛑 𝟐𝟐/𝟕 + (𝟐𝟐/𝟕)2

= 𝛑244/𝟕 π + (𝟐𝟐/𝟕)2

= 𝛑2𝟒𝟒𝛑/7 + 𝟒𝟖𝟒/𝟒𝟗

 

(xi) (2x+3) (2x+5)

Using (x + a) (x + b) = x2 + x (a + b) ab we get

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

(2x+3) (2x+5) = (2x)2 +2x (3 + 5) + 3.5

= 4x2 +16x +15

 

(xii) (3x – 3) (3x + 4)

Using (x + a) (x + b) = x2 + x (a + b) ab we get

x = 3x, a = –3 and b = 4

(3x – 3) (3x + 4) = (3x)2 + 3x [(–3)+(4)] + (-3)4

= 9x2 + 3x –12

= 9x2 + 3x –12


  1. Expand : 

(i) (x + 3 ) (x – 3)

Using (a + b) (a – b) = a2 – b2 we get

a = x, b = 3

(x + 3) (x – 3) = x2 – 32

= x2 – 9

 

(ii) (3x – 5y) (3x + 5y)

Using (a + b) (a – b) = a2 – b2 we get

a = 3x, b = 5y

(3x – 5y) (3x + 5y) = (3x)2 – (5y)2

= 9x2 – 25y2

 

(iii) (x/3+y/2)( x/3y/2)

Using (a + b) (a – b) = a2 – b2 we get

a = x/3, b = y/2

(x/3+ y/2)( x/3y/2)= (x/3 )2 – (x/3)2

= 𝐱𝟐/9-𝐲2/𝟒

 

(iv) (x2 + y2) (x2 – y2)

Using (a + b) (a – b) = a2 – b2 we get

a = x2, b = y2

(x2 + y2) (x2 – y2) = (x2)2 – (y2)2

= x4 – y4

 

(v) (a2 + 4b2) (a + 2b) (a – 2b)

Using (a + b) (a – b) = a2 – b2 for 2nd and 3rd term we get

(a2 + 4b2) (a + 2b) (a – 2b) = (a2 + 4b2) [a2 – (2b)2]

= (a2 + 4b2) (a2 – 4b2)

Using the above identity once again we get

= (a2)2 – (4b2)2

= a4 – 16b4

 

(vi) (x – 4) (x + 4) (x – 3) (x + 4)

Using (a + b) (a – b) = a2 – b2 we get

(x – 4) (x + 4) (x – 3) (x + 4) = (x2 – 42) (x2 – 32)

= (x2 – 16) (x2 – 9)

Using (a + b) (a – b) = x2 – x (a + b) + ab

= (x2)2 – x (16+9) +16×9

= x4 – 25x2 +144

 

(vii) (x – a) (x + a)(𝟏/𝐱𝟏/𝐚)( 𝟏/𝐱 +𝟏/𝐚 )

(x2 – a2) )(( /𝐱 )2– (1/a )2)

x2 x 𝟏/𝐱 2 – x2 x 𝟏/𝐚 2 – a2 x 𝟏/𝐱 2 + a2 x 𝟏/𝐚 2

1 – x2/a2 − a2/ x2 + 1

2 – 𝐱𝟐/a𝟐 − 𝐚𝟐 /x𝟐


  1. Simplify the following:

(i) (2x – 3y)2 + 12xy

= (2x)2 + (3y)2 – 2.2x.3y + 12xy

= 4x2 + 9y2 -12xy +12xy

= 4x2 + 9y2

 

(ii) (3m + 5n)2 – (2n)2

= (3m)2 + (5n)2 + 2.3m.5n – 4n2

= 9m2 + 25n2 + 30mn – 4n2

= 9m2 + 30 mn +21n2

 

(iii) (4a – 7b)2 – (3a)2

= (4a)2 – 2.4a.7b + (7b)2 – (3a)2

= 16a2 – 56ab + 49b2 – 9a2

= 7a2 -56ab + 49b2

 

(iv) (x + 𝟏/x)2 (m + 𝟏/𝐦 )2

= (x2 + 2. x. 𝟏/x + 𝟏/x2)2 – (m2 + 2. m. 𝟏/𝐦 + 𝟏/𝐦 2)2

= x2 + 2 + 𝟏/x 2 – (m2 + 2 + 𝟏/𝐦 2)

= x2 + 2 + 𝟏/x 2 – m2 + 2𝟏/𝐦 2

= x2 – m2+ 𝟏/x 𝟐𝟏/𝐦 𝟐 + 4

 

(v) (m2 + 2n2)2 – 4m2n2

= m4 +2m4.2n2 +4n4 – 4m2 n2

= m4+ 4m2n2 + 4n2 – 4m2n2

= m4 + 4n2

 

(vi) (3a – 2)2 – (2a -3)2

= (9a2 – 2.3a.2 + 22) – (4a2 – 2.3a.2 + 92)

= 9a2 –12a + 4 – 4a2 + 12a – 9

= 5a2 – 5

=5(a2 – 1)


Multiplication of Polynomials - Exercise 3.1.2-Class 9


 

One thought on “Multiplication of Polynomials – Exercise 3.1.2-Class 9

Comments are closed.