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}   ) 

Answer:

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 )

Answer:

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}   ) 

Answer:

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 ) 

Answer:

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 ) 

Answer:

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 )

Answer:

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}   

Answer:

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