Question 1: Find the following products:

i)  $\displaystyle ( y+9 ) ( y-9 )$          ii)  $\displaystyle ( 4+b ) ( 4-b )$

iii)  $\displaystyle ( 3x-5 ) ( 3x+5 )$          iv)  $\displaystyle ( a- \frac{2}{3} ) ( a+ \frac{2}{3} )$

i)  $\displaystyle ( y+9 ) ( y-9 ) = y^2-81$

ii)  $\displaystyle ( 4+b ) ( 4-b ) = 16-b^2$

iii)  $\displaystyle ( 3x-5 ) ( 3x+5 ) = 9x^2-25$

iv)  $\displaystyle ( a- \frac{2}{3} ) ( a+ \frac{2}{3} ) = a^2- \frac{4}{9}$

$\displaystyle \\$

Question 2: Find the following products:

i)  $\displaystyle ( 3x-5 ) ( 3x+5 )$          ii)  $\displaystyle ( 2+7x ) ( 2-7x )$

iii)  $\displaystyle ( \frac{a}{2} +3 ) ( \frac{a}{2} -3 )$         iv)  $\displaystyle ( 4x+3y ) ( 4x-3y )$

i)  $\displaystyle ( 3x-5 ) ( 3x+5 ) = 9x^2-25$

ii)  $\displaystyle ( 2+7x ) ( 2-7x ) = 4-49x^2$

iii)  $\displaystyle ( \frac{a}{2} +3 ) ( \frac{a}{2} -3 ) = \frac{a^2}{4} -9$

iv)  $\displaystyle ( 4x+3y ) ( 4x-3y ) = 16x^2-9y^2$

$\displaystyle \\$

Question 3: Find the following products:

i)  $\displaystyle ( \frac{a}{3} - \frac{b}{4} ) ( \frac{a}{3} + \frac{b}{4} )$           ii)  $\displaystyle ( \frac{t}{2} - \frac{1}{3} ) ( \frac{t}{2} + \frac{1}{3} )$

i)  $\displaystyle ( \frac{a}{3} - \frac{b}{4} ) ( \frac{a}{3} + \frac{b}{4} ) = \frac{a^2}{9} - \frac{b^2}{16}$

ii)  $\displaystyle ( \frac{t}{2} - \frac{1}{3} ) ( \frac{t}{2} + \frac{1}{3} ) = \frac{t^2}{4} - \frac{1}{9}$

$\displaystyle \\$

Question 4: Find the following products:

i)  $\displaystyle ( \frac{2}{x} + \frac{3}{y} ) ( \frac{2}{x} - \frac{3}{y} )$      ii)  $\displaystyle ( \frac{1}{a} - \frac{1}{b} ) ( \frac{1}{a} + \frac{1}{b} )$

iii)  $\displaystyle ( \frac{1}{3x} + \frac{2}{5y} ) ( \frac{1}{3x} - \frac{2}{5y} )$      iv)  $\displaystyle ( 1.1x-0.3y ) ( 1.1x+0.3y )$

i)  $\displaystyle ( \frac{2}{x} + \frac{3}{y} ) ( \frac{2}{x} - \frac{3}{y} ) = \frac{4}{x^2} - \frac{9}{y^2}$

ii)  $\displaystyle ( \frac{1}{a} - \frac{1}{b} ) ( \frac{1}{a} + \frac{1}{b} ) = \frac{1}{a^2} - \frac{1}{b^2}$

iii)  $\displaystyle ( \frac{1}{3x} + \frac{2}{5y} ) ( \frac{1}{3x} - \frac{2}{5y} ) = \frac{1}{9x^2} - \frac{2}{25y^2}$

iv)  $\displaystyle ( 1.1x-0.3y ) ( 1.1x+0.3y ) = 1.21x^2-0.09y^2$

$\displaystyle \\$

Question 5: Find the following products:

i)  $\displaystyle ( a^2+2b^2 ) ( a^2-2b^2 )$          ii)  $\displaystyle ( 6x^2-7y^2 ) ( 6x^2+7y^2 )$

iii)  $\displaystyle ( 4x^2+2yz ) ( 2x^2-yz )$          iv)  $\displaystyle ( ab- \frac{3}{2} cd ) ( 2ab+3cd )$

i)  $\displaystyle ( a^2+2b^2 ) ( a^2-2b^2 ) = a^4-4b^4$

ii)  $\displaystyle ( 6x^2-7y^2 ) ( 6x^2+7y^2 ) = 36x^4-49y^4$

iii)  $\displaystyle ( 4x^2+2yz ) ( 2x^2-yz ) = 8x^4+4x^2yz-4x^2yz-2y^2z^2 = 8x^4-2y^2z^2$

iv)  $\displaystyle ( ab- \frac{3}{2} cd ) ( 2ab+3cd ) = 2a^2b^2-3abcd+3abcd- \frac{9}{2} c^2d^2 = 2a^2b^2- \frac{9}{2} c^2d^2$

$\displaystyle \\$

Question 6: Find the following products:

i)  $\displaystyle ( 2x+3 ) ( 2x-3 ) ( 4x^2+9 )$          ii)  $\displaystyle ( x+2y ) ( x-2y ) ( x^2+4y^2 )$

iii)  $\displaystyle ( a+bc ) ( a-bc ) ( a^2+b^2c^2 )$          iv)  $\displaystyle ( \frac{2}{5} +x ) ( \frac{2}{5} -x ) ( \frac{4}{25} +x^2 )$

i)  $\displaystyle ( 2x+3 ) ( 2x-3 ) ( 4x^2+9 ) = \ ( 4x^2-9 ) ( 4x^2+9 ) = \ 16x^2-81$

ii)  $\displaystyle ( x+2y ) ( x-2y ) ( x^2+4y^2 ) = \ ( x^2-4y^2 ) ( x^2+4y^2 ) = \ x^4-16y^4$

iii)  $\displaystyle ( a+bc ) ( a-bc ) ( a^2+b^2c^2 ) = \ ( a^2-b^2c^2 ) ( a^2+b^2c^2 ) = {\ a}^4-b^4c^4$

iv)  $\displaystyle ( \frac{2}{5} +x ) ( \frac{2}{5} -x ) ( \frac{4}{25} +x^2 ) = \ ( \frac{4}{25} -x^2 ) ( \frac{4}{25} +x^2 ) = \frac{16}{625} -x^4$

$\displaystyle \\$

Question 7: Using the identity $\displaystyle ( a+b ) ( a-b ) = ( a^2-b^2 )$ , evaluate the following:

i)  $\displaystyle 88 \times 112$      ii)  $\displaystyle 153 \times 167$       iii)  $\displaystyle 10.8 \times 9.2$

iv)  $\displaystyle 3 \frac{1}{3} \times 4 \frac{2}{3}$       v)  $\displaystyle 9 \frac{1}{4} \times 15 \frac{3}{4}$

i)  $\displaystyle 88 \times 112 = ( 100-12 ) ( 100+12 ) = 1000-144 = 9856$
ii)  $\displaystyle 153 \times 167 = ( 160-7 ) ( 160+7 ) = 25600-49 = 25551$
iii)  $\displaystyle 10.8 \times 9.2 = ( 10+0.8 ) ( 10-0.8 ) = {10}^2-{0.8}^2 = 100-0.64 = 99.36$
iv)  $\displaystyle 3 \frac{1}{3} \times 4 \frac{2}{3} = ( 4- \frac{2}{3} ) ( 4+ \frac{2}{3} ) = 16- \frac{4}{9} = 15 \frac{5}{9}$
v)  $\displaystyle 9 \frac{1}{4} \times 15 \frac{3}{4} = ( \frac{25}{2} -3 \frac{1}{4} ) ( \frac{25}{2} +3 \frac{1}{4} ) = 145.6875$