Question

A- (1+7x^2)(1-7x^2)
B- (3x^2-y^2)(3x^2+y^2)
C- (x-2/3)(x+2/3)
D- (x/4+2/3)(x/4-2/3)

Answer 1

A- (1+7x^2)(1-7x^2)
=1-7x^2+7x^2-49x^4
=1-49x^4

B- (3x^2-y^2)(3x^2+y^2)
=9x^4+3x^2y^2-3x^2y^2-y^4
=9x^4-y^4

C- (x-2/3)(x+2/3)
=x^2/9+2x/9-2x/9-4/9
=x^2/9-4/9

D- (x/4+2/3)(x/4-2/3)
=(3x+8/12)(3x-8/12)
=9x^2-24x+24x-64/144
=9x^2-64/144

Answer 2

a)

$(1 + 7 {x}^{2})(1 - 7 {x}^{2} ) \\ = {1}^{2} - {(7 {x}^{2} )}^{2} = 1 - 49 {x}^{4}$

b)

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

c)

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

d)

$( \frac{x}{4} + \frac{2}{3})( \frac{x}{4} - \frac{2}{3} ) = {( \frac{x}{4})}^{2} -{( \frac{2}{3} )}^{2} \\ = \frac{ {x}^{2} }{16} - \frac{4}{9}$