Question

AJUTORRR!AM NEVOIE DE REZOLVAREA SUBPUNCTELOR a,b,c,d DAU COROANA

Answer 1

a) x² - 3 x - 4 = x² + x - 4 x - 4 = x ( x + 1 ) - 4 ( x + 1 ) = ( x + 1 ) ( x - 4 )

b) x² + 2 x - 3 = x² + 3 x - x - 3 = x ( x - 1 ) + 3 ( x - 1 ) = ( x - 1 ) ( x + 3 )

c) x² - x - 2 = x² + x - 2 x - 2 = x ( x + 1 ) - 2 ( x + 1 ) = ( x + 1 ) ( x - 2 )

d) x² + x - 6 = x² + x - 4 - 2 = x² - 4 + x - 2 = ( x - 2 ) ( x + 2 ) + ( x - 2) =

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

Answer 2

a)

x²-3x-4=

x²-4x+x-4=

x(x-4)-(x-4)-(x-4)(x-1)

b)

x²+2x-3=

x²-1x+3x-3=

x•(x-1)=

3•(x-1)=

(x+3)(x-1)

c)

x²-x-2=

x²-2x+1x-2=

x•(x-2)=

1•(x-2)=

(x+1)(x-2)

d)

x²+x-6=

x²+3x-2x-6=

x(x+3)-2(x+3)=

(x+3)(x-2)