Question

Descompuneti în factori aplicând formulă: a²-b²=(a+b) (a-b)


a) 4(x-1) ²-1 =

b) 9(2x+3) ²-x²=

c) 25x²-49(x+3) ²=

d) 25(2x-y)²-36(x+2y) ²=

e) x⁴-y⁴=

f) 16x⁴-81=

Answer 1

$a)4 {(x - 1)}^{2} - 1 = {2}^{2} {(x - 1)}^{2} - {1}^{2}$

$= {[2(x - 1)]}^{2} - {1}^{2}$

$= {(2x - 2)}^{2} - {1}^{2}$

$= (2x - 2 - 1)(2x - 2 + 1)$

$= (2x - 3)(2x - 1)$

$b)9 {(2x + 3)}^{2} - {x}^{2} = {3}^{2} {(2x + 3)}^{2} - {x}^{2}$

$= {[3(2x + 3)]}^{2} - {x}^{2}$

$= {(6x + 9)}^{2} - {x}^{2}$

$= (6x + 9 - x )(6x + 9 + x )$

$= (5x + 9)(7x + 9)$

$c)25 {x}^{2} - 49 {(x + 3)}^{2}$

$= {5}^{2} {x}^{2} - {7}^{2} {(x + 3)}^{2}$

$= {(5x)}^{2} - {[7(x + 3)]}^{2}$

$= {(5x)}^{2} - {(7x + 21)}^{2}$

$= [5x - (7x + 21)](5x + 7x + 21)$

$= (5x - 7x - 21)(12x + 21)$

$= ( - 2x - 21)(12x + 21)$

$d)25 {(2x - y)}^{2} - 36 {(x + 2y)}^{2}$

$= {5}^{2} {(2x - y)}^{2} - {6}^{2} {(x + 2y)}^{2}$

$= {[5(2x - y)]}^{2} - {[6(x + 2y)]}^{2}$

$= {(10x - 5y)}^{2} - {(6x + 12y)}^{2}$

$= [10x - 5y - (6x + 12y)](10x - 5y + 6x + 12y)$

$= (10x - 5y - 6x - 12y)(16x + 7y)$

$= (4x - 17y)(16x + 7y)$

$e) {x}^{4} - {y}^{4} = {( {x}^{2} )}^{2} - { ({y}^{2}) }^{2} = ( {x}^{2} - {y}^{2} )( {x}^{2} + {y}^{2} )$

$=(x-y)(x+y)( {x}^{2} + {y}^{2} )$

$f)16 {x}^{4} - 81 = {4}^{2} { ({x}^{2}) }^{2} - {9}^{2}$

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

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

$=[{(2x)}^{2}-{3}^{2}](4 {x}^{2} + 9)$

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