Question 1: Expand / Simplify the following:

(i) (3x+4y)^2

Answer:

(3x+4y)^2 = (3x)^2+2 \times 3x \times 4y + (4y)^2 = 9x^2 + 24xy + 16y^2

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(ii) (\sqrt{2}x-3y)^2

Answer:

(\sqrt{2}x-3y)^2 = (\sqrt{2}x)^2 + 2 \times (\sqrt{2}x) \times (-3y) + (-3y)^2 = 2x^2 - 6\sqrt{2}xy + 9y^2  

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(iii) \Big( 2x- \frac{1}{3x} \Big)^2

Answer:

\Big( 2x- \frac{1}{3x} \Big)^2 = (2x)^2 + 2 \times (2x) \times ( - \frac{1}{3x} ) + ( \frac{1}{3x})^2 = 4x^2- \frac{4}{3} + \frac{1}{9x^2}

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(iv) (x-1)(x+1)(x^2+1)(x^4+1)

Answer:

(x-1)(x+1)(x^2+1)(x^4+1)

= (x^2-1)(x^2+1)(x^4+1)

= (x^4-1)(x^4+1)

= (x^8-1)

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(v) \Big(x- \frac{1}{x} \Big) \Big(x+ \frac{1}{x} \Big) \Big( x^2+ \frac{1}{x^2} \Big) \Big( x^4+ \frac{1}{x^4} \Big)

Answer:

\Big(x- \frac{1}{x} \Big) \Big(x+ \frac{1}{x} \Big) \Big( x^2+ \frac{1}{x^2} \Big) \Big( x^4+ \frac{1}{x^4} \Big)

= \Big(x^2- \frac{1}{x^2} \Big)  \Big( x^2+ \frac{1}{x^2} \Big) \Big( x^4+ \frac{1}{x^4} \Big)

= \Big(x^4- \frac{1}{x^4} \Big)  \Big( x^4+ \frac{1}{x^4} \Big)

= \Big(x^8- \frac{1}{x^8} \Big)

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(vi) (2x- \frac{3}{x} +1)(2x+ \frac{3}{x} + 1)

Answer:

(2x- \frac{3}{x} +1)(2x+ \frac{3}{x} + 1)

= (2x+1 - \frac{3}{x} )(2x+1 + \frac{3}{x} )

= (2x+1)^2 - \Big( \frac{3}{x} \Big)^2

= 4x^2 + 4x + 1 - \frac{9}{x^2}

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(vii) (2x+5y+3)(2x+5y+4)

Answer:

Let 2x+5 = a

\Rightarrow  (2x+5y+3)(2x+5y+4)

= (a+3)(a+4) 

= a^2 + 7a + 12

= (2x+5)^2 + 7(2x+5)+ 12 

= 4x^2 + 20x + 25 + 14x + 35 + 12 

= 4x^2 + 34x + 72 

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(viii)  (1.5x^2-0.3y^2)(1.5x^2+0.3y^2)

Answer:

(1.5x^2-0.3y^2)(1.5x^2+0.3y^2) = (1.5x^2)^2 - (0.3y^2)^2 = 2.25x^4-0.09y^4

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(ix) (x^3-3x^2-x)(x^2-3x+1)

Answer:

(x^3-3x^2-x)(x^2-3x+1)

= x(x^2+3x+1)(x^2-3x+1)

= x \Big( (x^2+3x)^2 - 1 \Big)

= x(x^4-6x^3+9x^2-1)

= x^5-6x^4+9x^3-1

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(x) (2x^4-4x^2+1)(2x^4-4x^2-1)

Answer:

(2x^4-4x^2+1)(2x^4-4x^2-1)

= (2x^4-4x^2)^2 - 1

= 4x^8-16x^6+16x^4-1

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(xi) \Big(x+ \frac{2}{x} -3\Big) \Big(x- \frac{2}{x} -3 \Big)

Answer:

\Big(x+ \frac{2}{x} -3\Big) \Big(x- \frac{2}{x} -3 \Big)

= \Big( (x - 3 )+ \frac{2}{x} \Big) \Big( (x-3) - \frac{2}{x} \Big)

= x^2-6x+9 - \frac{4}{x^2}

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(xii) (5-2x)(5+2x)(25+4x^2)

Answer:

(5-2x)(5+2x)(25+4x^2)  = (25-4x^2)(25+4x^2)  = 625-16x^4

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(xiii) (x+2y+3)(x+2y+7)

Answer:

(x+2y+3)(x+2y+7)

= (x+2y)^2 +10(x+2y) + 21

= x^2+4xy+4y^2+10x+20y+21

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(xiv) (x+1)(x+2)(x+3)

Answer:

(x+1)(x+2)(x+3)

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

= x^3+3x^2+2x+3x^2+9x+6

= x^3 +6x^2 +11x +6

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(xv) (a+2b+c)^2

Note: We will use the following identify:    {(a+b+c)}^2=\left(a^2+b^2+c^2\right)+2(ab+bc+ca)   

Answer:

(a+2b+c)^2  = a^2 + 4b^2 +c^2 + 4ab + 4bc+ 2ca

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(xvi) \Big( \frac{x}{y} +  \frac{y}{z} +  \frac{z}{x} \Big)^2

Answer:

\Big( \frac{x}{y} +  \frac{y}{z} +  \frac{z}{x} \Big)^2

= (x^4+y^4+z^4 + 2x^2y^2-2y^2z^2-2z^2x^2) - (x^4+y^4+z^4 - 2x^2y^2-2y^2z^2+2z^2x^2)

= x^4+y^4+z^4 + 2x^2y^2-2y^2z^2-2z^2x^2 -x^4-y^4-z^4 + 2x^2y^2+2y^2z^2-2z^2x^2

= 4x^2y^2 - 4z^2x^2

= 4x^2 (y^2-z^2)

= 4x^2(y+z)(y-z)

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(xvii) (-2x+3y+2z)^2

Answer:

(-2x+3y+2z)^2 = 4x^2+9y^2+4z^2 -12xy+12yz-8zx

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(xviii) (a+b+c)^2 - (a-b+c)^2

Answer:

(a+b+c)^2 - (a-b+c)^2

= \Big(  a+b+c + a-b+c \Big)  \Big(  a+b+c -a +b -c \Big)

= (  2a+2c ) ( 2b ) = 4ab + 4bc

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(xix) (x^2+y^2+z^2)^2-(x^2-y^2+z^2)^2

Answer:

(x^2+y^2+z^2)^2-(x^2-y^2+z^2)^2

= \Big(  x^2+y^2+z^2 + x^2-y^2+z^2 \Big)  \Big(  x^2+y^2+z^2 - x^2 +y^2 - z^2 \Big)

= (2x^2+2z^2)(2y^2) = 4x^2y^2+4y^2z^2

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(xx) (x+y+z)^2+ \Big( x+ \frac{y}{2} +  \frac{z}{3} \Big)^2 - \Big( \frac{x}{2} + \frac{y}{3} + \frac{z}{4} \Big)^2

Answer:

(x+y+z)^2+ \Big( x+ \frac{y}{2} +  \frac{z}{3} \Big)^2 - \Big( \frac{x}{2} + \frac{y}{3} + \frac{z}{4} \Big)^2

= x^2+y^2+z^2 + 2xy + 2yz+2zx + x^2 + \frac{1}{4} y^2 + \frac{1}{9} z^2 + xy +  \frac{1}{3}  yz +

 \frac{2}{3} zx - \frac{1}{4} x^2 - \frac{1}{9} y^2 - \frac{1}{16} z^2 - \frac{1}{3} xy - \frac{1}{6} yz - \frac{1}{4} zx

= x^2( 1 + 1 - \frac{1}{4} ) + y^2(1+ \frac{1}{4} - \frac{1}{9}  ) +z^2(1 +  \frac{1}{9} -\frac{1}{16} )   \\ +xy(2+1- \frac{1}{3} )

+ yz(2 + \frac{1}{3} - \frac{1}{6} ) +zx(2+ \frac{2}{3} - \frac{1}{4} )

= \frac{7}{4} x^2 + \frac{41}{36} y^2 + \frac{151}{144} z^2 + \frac{8}{3} xy + \frac{13}{6} yz + \frac{29}{12} zx

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(xx) (x^2-x+1)^2-(x^2+x+1)^2

Answer:

(x^2-x+1)^2-(x^2+x+1)^2

= (x^2-x+1 + x^2+x+1)(x^2-x+1 - x^2-x-1)

= (2x^2+2)(-2x) = -4x^3-4x

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(xxi) \Big( x+ \frac{2}{x} \Big)^2 + \Big( x- \frac{2}{x} \Big)^2

Answer:

We will use the identity    {a^3+b^3= \ (a+b)}^3-3ab(a+b)   

\Rightarrow a = x+ \frac{2}{x}

\Rightarrow b = x- \frac{2}{x}

\Rightarrow a + b = 2x

\Big( x+ \frac{2}{x} \Big)^2 + \Big( x- \frac{2}{x} \Big)^2

= (2x)^3 - 3(x + \frac{2}{x} )( x- \frac{2}{x} ) (2x)

= 8x^3-6x^3+ \frac{24}{x} = 2x^3 + \frac{24}{x}

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(xxii) (2x-5y)^3-(2x+5y)^3

Answer:

We will use the identity    {a^3-b^3=\ (a-b)}^3+3ab(a-b)   

\Rightarrow a = 2x-5y \ and \ b = 2x+5y

\Rightarrow a - b = 2x-5y-2x-5y = -10y

Therefore (2x-5y)^3-(2x+5y)^3

= (-10y)^3 + 3 (2x-5y)(2x+5y)(-10y)

= -1000y^3 -120yx^2+750y^3

= -250y^3 -120yx^2

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(xxiii) (4x-5y)(16x^2+20xy+25y^2)

Answer:

We will use the identity a^3-b^3 = (a-b)(a^2 +ab + b^2)

(4x-5y)(16x^2+20xy+25y^2)

= (4x-5y)\Big[ (4x)^2 + (4x)(5y) + (5y)^2 \Big]

= (4x)^3 -(5y)^3 = 64x^3 - 125y^3

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(xxiv) (x^3+1)(x^6-x^3+1)

Answer:

We will use the identity a^3-b^3 = (a-b)(a^2 +ab + b^2)

(x^3+1)(x^6-x^3+1)

= (x^3+1) \Big[ (x^3)^2 -(x^3)(1) +(1)^2 \Big]

= (x^3)^3 +(1)^3 = x^9 +1

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(xxv) (4x-3y+2z)(16x^2+9y^2+4z^2+12xy+6yz-8zx) 

Answer:

We will use the identity a^3+b^3+c^3 -3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)   

(4x-3y+2z)(16x^2+9y^2+4z^2+12xy+6yz-8zx) 

= (4x-3y+2z) \Big[ (4x)^2 +(-3y)^2 + (2z)^2 - (4x)(-3y) - (-3y)(2z) - (2z)(4x) \Big] 

= (4x)^3+(-3y)^3+(2z)^3 - 3(4x)(-3y)(2z) 

= 64x^3 - 27y^3+8z^3 +72xyz 

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Question 2: Evaluate the following identities:

(i) 103 \times 97

Answer:

103 \times 97  = (100+3)(100-3) = 100^2 - 3^2 = 10000-9 = 9991

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(ii) (97)^2

Answer:

(97)^2  = (100-3)^2 = 10000 -600 +9 = 9409

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(iii) 0.54  \times 0.54 -0.46  \times 0.46

Answer:

0.54  \times 0.54 -0.46  \times 0.46 = 0.54^2-0.46^2 = (0.54+0.46)(0.54-0.46) = 1 \times 0.08 = 0.08

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(iv) (0.98)^2

Answer:

(0.98)^2  = (1-0.02)^2 = 1 -0.04+0.0004 = 0.9604

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(v) 991 \times 1009

Answer:

991 \times 1009  = (1000 -9) (1000 + 9) = 1000^2 -9^2 = 1000000-81 = 999919

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(vi) (1002)^3

Answer:

We will use the identity    {(a+b)}^3=\ a^3+b^3+3ab(a+b)   

(1002)^3 = (1000+2)^3 = 1000^3 + 2^3 + 3\times 1000 \times 2 (1000 + 2)

= 1000000000+8 + 6012000= 1006012008

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(vii) (999)^3

Answer:

We will use the identity    {(a-b)}^3=\ a^3-b^3-3ab(a-b)   

(999)^3  = (1000 - 1)^3 = 1000^3 -1^3-3 \times 1000 \times 1 (1000-1)

= 1000000000-1-299700 = 997002999

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(viii) (103)^3

Answer:

We will use the identity    {(a+b)}^3=\ a^3+b^3+3ab(a+b)   

(103)^3  = (100 + 3)^3 = 100^3 + 3^3 + 3 \times 100 \times 3 (100+3)

= 1000000+27+92700 = 1092727

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(ix) (10.4)^3

Answer:

(10.4)^3  = (10+0.4)^3 = 10^3 +0.4^3 + 3 \times 10 \times 0.4 (10+0.4)

= 1000 + 0.064+124.8 = 1124.864

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(x) (99)^3

Answer:

We will use the identity    {(a-b)}^3=\ a^3-b^3-3ab(a-b)   

(99)^3  = (100 -1 )^3 = 100^3 - 1^3 - 3 \times 100 \times 1 (100 -1 )

= 1000000-1-29700=970299

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(xi) 111^3 - 89^3

Answer:

We will use the identity     {a^3-b^3=\ (a-b)}^3+3ab(a-b)   

111^3 - 89^3 = (100+11)^3 - (100-11)^3

= (100+11-100+11)^3 + 3 (100+11)(100-11) (22)

=22^3 + 3 \times 111 \times 89 \times 22

= 662662

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(xii) 104^3+96^3

Answer:

We will use the identity  a^3+b^3= (a+b)^3-3ab(a+b)

104^3+96^3

= (100+4)^3 + (100-4)^3

= (100+4+100-4) - \Big[ (100+4+100-4)^3 - 3 (100+4)(100-4)(100+4+100-4) \Big]

= 200^3 - 3 \times 104 \times 96 \times 200

= 2009600

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(xiii) \Big( \frac{1}{2} \Big)^3 + \Big( \frac{1}{3} \Big)^3 - \Big( \frac{5}{6} \Big)^3

Answer:

We are going to use this identity a^3+b^3+c^3 -3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)   

\Big( \frac{1}{2} \Big)^3 + \Big( \frac{1}{3} \Big)^3 - \Big( \frac{5}{6} \Big)^3

= ( \frac{1}{2} + \frac{1}{3} - \frac{5}{6} ) \Big ( ( \frac{1}{2})^2 + ( \frac{1}{3} )^2  + (- \frac{5}{6} )^2 - \frac{1}{2} \times \frac{1}{3} + \frac{1}{3} \times \frac{5}{6} + \frac{5}{6} \times \frac{1}{2} \Big) + 3 \times \frac{1}{2} \times \frac{1}{3} \times (- \frac{5}{6} ) 

= \frac{0}{6} \Big ( ( \frac{1}{2})^2 +( \frac{1}{3} )^2 + (- \frac{5}{6})^2 - \frac{1}{2} \times \frac{1}{3} + \frac{1}{3} \times \frac{5}{6} + \frac{5}{6} \times \frac{1}{2} \Big) - \frac{5}{12}

= \frac{-5}{12}

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Question 3:

(i) If the number a is 7 more than number b and the sum of the squares of a and b is 85 , find the product of ab

Answer:

a+b = 7 \Rightarrow a - b = 7

a^2 + b^2 = 85

Therefore (a-b)^2 + 2ab = 85

\Rightarrow 7^2 + 2ab = 85

\Rightarrow 2ab = 85 - 49 = 36

\Rightarrow ab = 18

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(ii) If the number x is 3  less than the number y and the sum if the square of x and y is 29 , find xy

Answer:

x+3 = 7 \Rightarrow x - y = -3

x^2 + y^2 = 29

(x-y)^2 +2xy = 29

\Rightarrow (-3)^2 + 2xy = 29

\Rightarrow 2xy = 29 - 9 = 20

\Rightarrow xy = 10

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(iii) If the sum of two numbers is 7 and the sum of their cubes is 133 , find the sum of their squares.

Answer:

a+ b = 7

a^3 + b^3 = 133

(a+b)^3 = a^3 +b^3 +3ab(a+b)

7^2 = 133 + 3ab (7)

21ab = 133 - 49 = 84

\Rightarrow ab = 4

Therefore (a+b)^2 -2ab = a^2 + b^2

\Rightarrow 7^2 - 8 = a^2 +b^2 

\Rightarrow a^2 +b^2 = 41

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Exercise 5(a) continued…

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