Question

Descompuneti in produs de factori:
a) (x²+x)(x²+x-8)+12
b)x(x+2)(x²+2x-14)+48
c)x(x+4)(x²+4x+6)+8
d) (x²+4x+1)(x²+4x-2)-18

Va rog frumos!

Answer 1

a) notezi x²+2xsi inlocuiesti in exercitiu ⇒y(y-8)+12=y²-8y+12=y²-6y-2y+12=y(y-6)-2(y-6)=(y-6)(y-2)
revenind la notatie (x²+x-6)(x²+x-2)=(x²+3x-2x-6)(x²+2x-x-2)=(x(x+3)-2(x+3))(x(x+2)-(x+2))=(x+3)(x-2)(x+2)(x-1)
b)(x²+2x)(x²+2x-14)+48
notezi x²+2x rezulta y(y-14)+48=y²-14y+48=y²-6y-8y+48=y(y-6)-8(y-6)=(y-6)(y-8)
(x²+2x-6)(x²+2x-8)
celelalte exercitii le faci la fel
SUCCES!

Answer 2

a)
Notam x^2+x=y

(x²+x)(x²+x-8)+12=y(y-8)+12=y^2-8y+12=(y-6)(y-2)
y-6=x^2+x-6=(x-2)(x+3)
y-2=x^2+x-2=(x-1)(x+2)

(x²+x)(x²+x-8)+12=(x-2)(x+3)(x-1)(x+2)=(x^2 -4)(x+3)(x-1)

b)
Notam y=x^2+2x

x(x+2)(x²+2x-14)+48=y(y-14)+48=y^2-14y+48=(y-6)(y-8)
y-6=x^2+2x-6
y-8=x^2+2x-8=(x-2)(x+4)

x(x+2)(x²+2x-14)+48=(x^2+2x-6)(x-2)(x+4)

c)
Notam y=x^2+4x

x(x+4)(x²+4x+6)+8=y(y+6)+8=y^2+6y+8=(y+2)(y+4)
y+2=x^2+4x+2
y+4=x^2+4x+4=(x+2)^2

x(x+4)(x²+4x+6)+8=(x^2+4x+2)(x+2)^2

d)
Notam y=x^2+4x

 (x²+4x+1)(x²+4x-2)-18=(y+1)(y-2)-18=y^2-y-2-18=y^2-y-20=(y+4)(y-5)
y+4=x^2+4x+4=(x+2)^2
y-5=x^2+4x-5=(x-1)(x+5)

(x²+4x+1)(x²+4x-2)-18=(x-1)(x+5)(x+2)^2