Class 8: Special Products and Expansions – Exercise 19A – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Find the following products;

$\displaystyle \text{i) } ( x+5 ) ( x+7 )$ $\displaystyle \text{ii) } ( b+2 ) ( b+9 )$ $\displaystyle \text{iii) } ( c+2 ) ( c+ \frac{3}{5} )$ $\displaystyle \text{iv) } ( t+ \frac{4}{3} ) ( t+ \frac{1}{3} )$

Answer:

$\displaystyle \text{i) } ( x+5 ) ( x+7 ) = x^2+5x+7x+35 = x^2+12x+35$

$\displaystyle \text{ii) } ( b+2 ) ( b+9 ) = b^2+2b+9b+18 = b^2+11b+18$

$\displaystyle \text{iii) } ( c+2 ) ( c+ \frac{3}{5} ) = c^2+2c+ \frac{3}{5} c+ \frac{6}{5} = c^2+ ( 2+ \frac{3}{5} ) c+ \frac{6}{5} = c^2+ \frac{13}{5} c+ \frac{6}{5}$

$\displaystyle \text{iv) } ( t+ \frac{4}{3} ) ( t+ \frac{1}{3} ) = { t}^2+ \frac{4}{3} t+ \frac{1}{3} t+ \frac{4}{9} = { t}^2+ \frac{5}{3} t+ \frac{4}{9}$

$\displaystyle \\$

Question 2: Find the following products:

$\displaystyle \text{i) } ( y+8 ) ( y-4 )$ $\displaystyle \text{ii) } ( z+6 ) ( z-11 )$ $\displaystyle \text{iii) } ( c-5 ) ( c+1 )$ $\displaystyle \text{iv) } ( b-13 ) ( b+10 )$

Answer:

$\displaystyle \text{i) } ( y+8 ) ( y-4 ) = y^2+8y-4y-32 = { y}^2+4y-32$

$\displaystyle \text{ii) } ( z+6 ) ( z-11 ) = { z}^2+6z-11z-66 = { z}^2-5z-66$

$\displaystyle \text{iii) } ( c-5 ) ( c+1 ) = { c}^2-5c+c-5 = c^2-4c-5$

$\displaystyle \text{iv) } ( b-13 ) ( b+10 ) = { b}^2-13b+10b-130 = b^2-3b-130$

$\displaystyle \\$

Question 3: Find the following products:

$\displaystyle \text{i) } ( x-3 ) ( x-6 )$ $\displaystyle \text{ii) } ( z-11 ) ( z-4 )$ $\displaystyle \text{iii) } ( b-6 ) ( b-8 )$ $\displaystyle \text{iv) } ( a- \frac{3}{5} ) ( a- \frac{1}{3} )$

Answer:

$\displaystyle \text{i) } ( x-3 ) ( x-6 ) = { x}^2-3x-6x+18 = { x}^2-9x+18$

$\displaystyle \text{ii) } ( z-11 ) ( z-4 ) = z^2-15z+44 = z^2-11z-4z+44$

$\displaystyle \text{iii) } ( b-6 ) ( b-8 ) = b^2-6b-8b+48 = { b}^2-14b+48$

$\displaystyle \text{iv) } ( a- \frac{3}{5} ) ( a- \frac{1}{3} ) = { a}^2- \frac{3}{5} a- \frac{1}{3} a+ \frac{1}{5} = a^2- \frac{14}{15} a+ \frac{1}{5}$

$\displaystyle \\$

Question 4: Explain the following:

$\displaystyle \text{i) } ( x- \frac{1}{2} ) ( x+ \frac{3}{2} )$ $\displaystyle \text{ii) } ( p-3 ) ( p+ \frac{1}{2} )$

Answer:

$\displaystyle \text{i) } ( x- \frac{1}{2} ) ( x+ \frac{3}{2} ) = { x}^2- \frac{1}{2} x+ \frac{3}{2} x- \frac{3}{4} = { x}^2+x- \frac{3}{4}$

$\displaystyle \text{ii) } ( p-3 ) ( p+ \frac{1}{2} ) = { p}^2-3p- \frac{1}{2} p- \frac{3}{2} = { p}^2- \frac{7}{2} p- \frac{3}{2}$

$\displaystyle \\$

Question 5: Explain the following:

$\displaystyle \text{i) } ( 4p+3 ) ( 4p+7 )$ $\displaystyle \text{ii) } ( 9c+4 ) ( 9c-2 )$ $\displaystyle \text{ii) } ( 3a-8 ) ( 3a+2 )$

$\displaystyle \text{iv) } ( 5x-2 ) ( 5x-7 )$

Answer:

$\displaystyle \text{i) } ( 4p+3 ) ( 4p+7 ) = { 16p}^2+12p+28p+21 = { 16p}^2+40p+21$

$\displaystyle \text{ii) } ( 9c+4 ) ( 9c-2 ) = 81c^2+36c-18c-8 = 81c^2+18c-8$

$\displaystyle \text{ii) } ( 3a-8 ) ( 3a+2 ) = {9a}^2-24a+6a-16 = { 9a}^2-18a-16$

$\displaystyle \text{iv) } ( 5x-2 ) ( 5x-7 ) = {25x}^2-10x-35x+14 = { 25x}^2-45x+14$

$\displaystyle \\$

Question 6: Find the following products:

$\displaystyle \text{i) } ( x^2+3 ) ( x^2+6 )$ $\displaystyle \text{ii) } ( y^2-1 ) ( y^2+4 )$

$\displaystyle \text{iii) } ( z^2+3 ) ( z^2-7 )$ $\displaystyle \text{iv) } ( t^2-2 ) ( t^2-5 )$

Answer:

$\displaystyle \text{i) } ( x^2+3 ) ( x^2+6 ) = { x}^4+{3x}^2+6x^2+18 = x^4+9x^2+18$

$\displaystyle \text{ii) } ( y^2-1 ) ( y^2+4 ) = y^4-y^2+{4y}^2-4 = { y}^4+3y^2-4$

$\displaystyle \text{iii) } ( z^2+3 ) ( z^2-7 ) = { z}^4+3z^2-7z^2-21 = { z}^4-4z^2-21$

$\displaystyle \text{iv) } ( t^2-2 ) ( t^2-5 ) = { t}^4-2t^2-5t^2+10 = t^4-7t^2+10$

$\displaystyle \\$

Question 7: Find the following products:

$\displaystyle \text{i) } ( ab-2 ) ( ab+4 )$ $\displaystyle \text{ii) } ( 2+xy ) ( 3-xy )$

Answer:

$\displaystyle \text{i) } ( ab-2 ) ( ab+4 ) = { a}^2b^2-2ab++4ab-8 = { a}^2b^2+2ab-8$

$\displaystyle \text{ii) } ( 2+xy ) ( 3-xy ) = {6+3xy-2xy-x}^2y^2 = { 6+xy-x}^2y^2$

$\displaystyle \\$

Question 8: Find the following products:

$\displaystyle \text{i) } ( 3x+4y ) ( 4x+3y )$ $\displaystyle \text{ii) } ( 4a-5b ) ( 3a+2b )$

$\displaystyle \text{iii) } ( 2y+z ) ( 7z-3y )$ $\displaystyle \text{iv) } ( 2m-4n ) ( 4m-3n )$

Answer:

$\displaystyle \text{i) } ( 3x+4y ) ( 4x+3y ) = 12x^2+16xy+9xy+12y^2 = 12x^2+25xy+12y^2$

$\displaystyle \text{ii) } ( 4a-5b ) ( 3a+2b ) = 12a^2-15ab+8ab-10b^2 = 12a^2-7ab-10b^2$

$\displaystyle \text{iii) } ( 2y+z ) ( 7z-3y ) = 14yz+7z^2-6y^2-3yz = 7z^2+11yz-6y^2$

$\displaystyle \text{iv) } ( 2m-4n ) ( 4m-3n ) = 8m^2-16mn-9mn-12n^2 = { m}^2-25mn-12n^2$

$\displaystyle \\$

Question 9: Find the following products:

$\displaystyle \text{i) } ( 2pq+0.1mn ) ( 0.2pq+3mn )$ $\displaystyle \text{ii) } ( \frac{4m}{p} - \frac{0.2n}{q} ) ( \frac{3m}{p} + \frac{0.5}{q} )$

$\displaystyle \text{iii) } ( \frac{3}{4} x-2pq ) ( 3x- \frac{4}{5} pq )$

Answer:

$\displaystyle \text{i) } ( 2pq+0.1mn ) ( 0.2pq+3mn )$

$\displaystyle = 0.4p^2q^2+0.02pqmn+6pqmn+0.3m^2n^2$

$\displaystyle = 0.4p^2q^2+6.02pqmn+0.3m^2n^2$

$\displaystyle \text{ii) } ( \frac{4m}{p} - \frac{0.2n}{q} ) ( \frac{3m}{p} + \frac{0.5}{q} )$

$\displaystyle = \frac{12m^2}{p^2} - \frac{0.6mn}{pq} + \frac{2mn}{pq} - \frac{0.1n^2}{q^2}$

$\displaystyle = \frac{12m^2}{p^2} + \frac{1.4mn}{pq} - \frac{0.1n^2}{q^2}$

$\displaystyle \text{iii) } ( \frac{3}{4} x-2pq ) ( 3x- \frac{4}{5} pq )$

$\displaystyle = \frac{9}{4} x^2-6pqx- \frac{3}{5} pqx- \frac{8}{5} p^2q^2$

$\displaystyle = \frac{9}{4} x^2- \frac{33}{5} pqx- \frac{8}{5} p^2q^2$

$\displaystyle \\$

Question 10: Find the following products:

$\displaystyle \text{i) } ( 2a^2+3b^2 ) ( 3a^2+2b^2 )$ $\displaystyle \text{ii) } ( x^2+5y^2 ) ( y^2-2x^2 )$

$\displaystyle \text{iii) } ( 3c^2-4d^2 ) ( 4c^2-3d^2 )$

Answer:

$\displaystyle \text{i) } ( 2a^2+3b^2 ) ( 3a^2+2b^2 ) = 4a^4+9a^2b^2+4a^2b^2+9a^2b^2 = 4a^4+13a^2b^2+9a^2b^2$

$\displaystyle \text{ii) } ( x^2+5y^2 ) ( y^2-2x^2 ) = { x}^2y^2+5y^4-2x^4-10x^2y^2 = 5y^4-9x^2y^2-2x^4$

$\displaystyle \text{iii) } ( 3c^2-4d^2 ) ( 4c^2-3d^2 ) = 12c^4-16c^2d^2-9c^2d^2+12d^4 = 12c^4-25c^2d^2+12d^4$

ICSE / ISC / CBSE Mathematics Portal for K12 Students

Mathematics Made Easy for School Students

Class 8: Special Products and Expansions – Exercise 19A

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