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Find the following products:
(i) $(x + 4) (x + 7)$
(ii) $(x - 11) (x + 4)$
(iii) $(x + 7) (x - 5)$
(iv) $(x - 3) (x - 2)$
(v) $(y^2 - 4) (y^2 - 3)$
(vi) $(x + \frac{4}{3}) (x + \frac{3}{4})$
(vii) $(3x + 5) (3x + 11)$
(viii) $(2x^2 - 3) (2x^2 + 5)$
(ix) $(z^2 + 2) (z^2 - 3)$
(x) $(3x - 4y) (2x - 4y)$
(xi) $(3x^2 - 4xy) (3x^2 - 3xy)$
(xii) $(x + \frac{1}{5}) (x + 5)$
(xiii) $(z + \frac{3}{4}) (z + \frac{4}{3})$
(xiv) $(x^2 + 4) (x^2 + 9)$
(xv) $(y^2 + 12) (y^2 + 6)$
(xvi) $(y^2 + \frac{5}{7}) (y^2 - \frac{14}{5})$
(xvii) $(p^2 + 16) (p^2 - \frac{1}{4})$
To do:
We have to find the given products.
Solution:
Here, to find the given products we can use distributive property twice.
Distributive Property:
The distributive property of multiplication states that when a factor is multiplied by the sum or difference of two terms, it is essential to multiply each of the two numbers by the factor, and finally perform the addition or subtraction operation.
$(a+b)(c+d)=a(c+d)+b(c+d)$..............(I)
Therefore,
(i) The given expression is $(x + 4) (x + 7)$.
$(x + 4) (x + 7)=x(x+7)+4(x+7)$ [Using (I)]
$(x + 4) (x + 7)=x(x)+x(7)+4(x)+4(7)$
$(x + 4) (x + 7)=x^2+7x+4x+28$
$(x + 4) (x + 7)=x^2+11x+28$
(ii) The given expression is $(x - 11) (x + 4)$
$(x - 11) (x + 4)=x(x+4)-11(x+4)$ [Using (I)]
$(x - 11) (x + 4)=x(x)+x(4)-11(x)-11(4)$
$(x - 11) (x + 4)=x^2+4x-11x-44$
$(x - 11) (x + 4)=x^2-7x-44$
(iii) The given expression is $(x + 7) (x - 5)$
$(x + 7) (x - 5)=x(x-5)+7(x-5)$ [Using (I)]
$(x + 7) (x - 5)=x(x)-x(5)+7(x)-7(5)$
$(x + 7) (x - 5)=x^2-5x+7x-35$
$(x + 7) (x - 5)=x^2+2x-35$
(iv) The given expression is $(x - 3) (x - 2)$
$(x - 3) (x - 2)=x(x-2)-3(x-2)$ [Using (I)]
$(x - 3) (x - 2)=x(x)-x(2)-3(x)+3(2)$
$(x - 3) (x - 2)=x^2-2x-3x+6$
$(x - 3) (x - 2)=x^2-5x+6$
(v) The given expression is $(y^2 - 4) (y^2 - 3)$
$(y^2 - 4) (y^2 - 3)=y^2(y^2-3)-4(y^2-3)$ [Using (I)]
$(y^2 - 4) (y^2 - 3)=y^2(y^2)-y^2(3)-4(y^2)+4(3)$
$(y^2 - 4) (y^2 - 3)=y^4-3y^2-4y^2+12$
$(y^2 - 4) (y^2 - 3)=y^4-7y^2+12$
(vi) The given expression is $(x+\frac{4}{3})(x + \frac{3}{4})$
$(x+\frac{4}{3})(x + \frac{3}{4})=x (x + \frac{3}{4})+\frac{4}{3}(x + \frac{3}{4})$ [Using (I)]
$(x+\frac{4}{3})(x+\frac{3}{4})=x(x)+x(\frac{3}{4})+\frac{4}{3}(x)+\frac{3}{4}(\frac{3}{4})$
$(x+\frac{4}{3})(x+\frac{3}{4})=x^2+x(\frac{3}{4}+\frac{4}{3})+1$
$(x+\frac{4}{3})(x+\frac{3}{4})=x^2+x(\frac{3\times3+4\times4}{12})+1$
$(x+\frac{4}{3})(x+\frac{3}{4})=x^2+x(\frac{9+16}{12})+1$
$(x+\frac{4}{3})(x+\frac{3}{4})=x^2+\frac{25}{12}x+1$
(vii) The given expression is $(3x + 5) (3x + 11)$
$(3x + 5) (3x + 11)=3x(3x+11)+5(3x+11)$ [Using (I)]
$(3x + 5) (3x + 11)=3x(3x)+3x(11)+5(3x)+5(11)$
$(3x + 5) (3x + 11)=9x^2+33x+15x+55$
$(3x + 5) (3x + 11)=9x^2+48x+55$
(viii) The given expression is $(2x^2 - 3) (2x^2 + 5)$
$(2x^2 - 3) (2x^2 + 5)=2x^2(2x^2+5)-3(2x^2+5)$ [Using (I)]
$(2x^2 - 3) (2x^2 + 5)=2x^2(2x^2)+2x^2(5)-3(2x^2)-3(5)$
$(2x^2 - 3) (2x^2 + 5)=4x^4+10x^2-6x^2-15$
$(2x^2 - 3) (2x^2 + 5)=4x^4+4x^2-15$
(ix) The given expression is $(z^2 + 2) (z^2 - 3)$
$(z^2 + 2) (z^2 - 3)=z^2(z^2-3)+2(z^2-3)$ [Using (I)]
$(z^2 + 2) (z^2 - 3)=z^2(z^2)-z^2(3)+2(z^2)-2(3)$
$(z^2 + 2) (z^2 - 3)=z^4-3z^2+2z^2-6$
$(z^2 + 2) (z^2 - 3)=z^4-z^2-6$
(x) The given expression is $(3x - 4y) (2x - 4y)$
$(3x - 4y) (2x - 4y)=3x(2x-4y)-4y(2x-4y)$ [Using (I)]
$(3x - 4y) (2x - 4y)=3x(2x)-3x(4y)-4y(2x)+4y(4y)$
$(3x - 4y) (2x - 4y)=6x^2-12xy-8xy+16y^2$
$(3x - 4y) (2x - 4y)=6x^2-20xy+16y^2$
(xi) The given expression is $(3x^2 - 4xy) (3x^2 - 3xy)$
$(3x^2 - 4xy) (3x^2 - 3xy)=3x^2(3x^2-3xy)-4xy(3x^2-3xy)$ [Using (I)]
$(3x^2 - 4xy) (3x^2 - 3xy)=3x^2(3x^2)-3x^2(3xy)-4xy(3x^2)+4xy(3xy)$
$(3x^2 - 4xy) (3x^2 - 3xy)=9x^4-9x^3y-12x^3y+12x^2y^2$
$(3x^2 - 4xy) (3x^2 - 3xy)=9x^4-21x^3y+12x^2y^2$
(xii) The given expression is $(x+\frac{1}{5})(x + 5)$
$(x+\frac{1}{5})(x + 5)=x (x + 5)+\frac{1}{5}(x + 5)$ [Using (I)]
$(x+\frac{1}{5})(x + 5)=x (x) + x(5)+\frac{1}{5}(x) + \frac{1}{5}(5)$
$(x+\frac{1}{5})(x + 5)=x^2 + x(5+\frac{1}{5}) + 1$
$(x+\frac{1}{5})(x + 5)=x^2 + x(\frac{5\times5+1}{5}) + 1$
$(x+\frac{1}{5})(x + 5)=x^2 + x(\frac{25+1}{5}) + 1$
$(x+\frac{1}{5})(x + 5)=x^2 + \frac{26}{5}x + 1$
(xiii) The given expression is $(z+\frac{3}{4})(z + \frac{4}{3})$
$(z+\frac{3}{4})(z + \frac{4}{3})=z (z + \frac{4}{3})+\frac{3}{4}(z + \frac{4}{3})$ [Using (I)]
$(z+\frac{3}{4})(z+\frac{4}{3})=z(z)+z(\frac{4}{3})+\frac{3}{4}(z)+\frac{3}{4}(\frac{4}{3})$
$(z+\frac{3}{4})(z+\frac{4}{3})=z^2+z(\frac{4}{3}+\frac{3}{4})+1$
$(z+\frac{3}{4})(z+\frac{4}{3})=z^2+z(\frac{4\times4+3\times3}{12})+1$
$(z+\frac{3}{4})(z+\frac{4}{3})=z^2+z(\frac{16+9}{12})+1$
$(z+\frac{3}{4})(z+\frac{4}{3})=z^2+\frac{25}{12}z+1$
(xiv) The given expression is $(x^2 + 4) (x^2 + 9)$
$(x^2 + 4) (x^2 + 9)=x^2(x^2+9)+4(x^2+9)$ [Using (I)]
$(x^2 + 4) (x^2 + 9)=x^2(x^2)+x^2(9)+4(x^2)+4(9)$
$(x^2 + 4) (x^2 + 9)=x^4+9x^2+4x^2+36$
$(x^2 + 4) (x^2 + 9)=x^4+13x^2+36$
(xv) The given expression is $(y^2 + 12) (y^2 + 6)$
$(y^2 + 12) (y^2 + 6)=y^2(y^2+6)+12(y^2+6)$ [Using (I)]
$(y^2 + 12) (y^2 + 6)=y^2(y^2)+y^2(6)+12(y^2)+12(6)$
$(y^2 + 12) (y^2 + 6)=y^4+6y^2+12y^2+72$
$(y^2 + 12) (y^2 + 6)=y^4+18y^2+72$
(xvi) The given expression is $(y^2+\frac{5}{7})(y^2-\frac{14}{5})$
$(y^2+\frac{5}{7})(y^2-\frac{14}{5})=y^2(y^2-\frac{14}{5})+\frac{5}{7}(y^2-\frac{14}{5})$ [Using (I)]
$(y^2+\frac{5}{7})(y^2-\frac{14}{5})=y^2(y^2)-y^2(\frac{14}{5})+\frac{5}{7}(y^2)-\frac{5}{7}(\frac{14}{5})$
$(y^2+\frac{5}{7})(y^2-\frac{14}{5})=y^4+y^2(\frac{-14}{5}+\frac{5}{7})-2$
$(y^2+\frac{5}{7})(y^2-\frac{14}{5})=y^4+y^2(\frac{-14\times7+5\times5}{35})-2$
$(y^2+\frac{5}{7})(y^2-\frac{14}{5})=y^4+y^2(\frac{-98+25}{35})-2$
$(y^2+\frac{5}{7})(y^2-\frac{14}{5})=y^4-\frac{73}{35}y^2-2$
(xvii) The given expression is $(p^2 + 16) (p^2 -\frac{1}{4})$
$(p^2 + 16) (p^2 -\frac{1}{4})=p^2(p^2-\frac{1}{4})+16(p^2-\frac{1}{4})$ [Using (I)]
$(p^2+16)(p^2-\frac{1}{4})=p^2(p^2)-p^2(\frac{1}{4})+16(p^2)-16(\frac{1}{4})$
$(p^2+16)(p^2-\frac{1}{4})=p^4+p^2(16-\frac{1}{4})-4$
$(p^2+16)(p^2-\frac{1}{4})=p^4+p^2(\frac{16\times4-1}{4})-4$
$(p^2+16)(p^2-\frac{1}{4})=p^4+p^2(\frac{64-1}{4})-4$
$(p^2+16)(p^2-\frac{1}{4})=p^4+\frac{63}{4}p^2-4$
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