Question

descompuneti in factori utilizând reguli de calcul in R:
a.) (x+1)pătrat - (x+1) - 2 =
b.) 3(x+1)pătrat+5(x+1)+2=
c.) (x pătrat + 3x +1)pătrat - 1=
d.) (xpatrat + x + 1)(xpatrat + x -3)+3=

Answer 1

[tex]a) (x+1)^2-(x+1)-2=\\ Not\breve{a}m\ x+1=t\\ t^2-t-2=t^2-2t+t-2=t(t-2)+(t-2)=(t-2)(t+1)\\ \hat{I}nlocuind: (x+1-2)(x+1+2)=\boxed{(x-1)(x+3)}\\ \\ b)3(x+1)^2+5(x+1)+2=\\ Se\ face\ la\ fel\ ca\ a).\\ 3t^2+5t+2=3t^2+3t+2t+2=3t(t+1)+2(t+1)=\\ =(t+1)(3t+2)\\ \hat{I}nlocuind:(x+1+1)[3(x+1)+2]=\boxed{(x+2)(3x+5)}\\ \\ c)(x^2+3x+1)^2-1=(x^2+3x+1-1)(x^2+3x+1+1)=\\ (x^2+3x)(x^2+3x+2)=x(x+3)(x^2+2x+x+2)=\\ x(x+3)[x(x+2)+(x+2)]=\boxed{x(x+1)(x+2)(x+3)}\\ [/tex]
[tex]d)(x^2+x+1)(x^2+x-3)+3=\\ Notam: x^2+x=t\\ (t+1)(t-3)+3=t^2-3t+t-3+3=t^2-2t=t(t-2)\\ \hat{I}nlocuind: (x^2+x)(x^2+x-2)=x(x+1)(x^2+2x-x-2)=\\ =x(x+1)[x(x+2)-(x+2)]=\boxed{x(x+1)(x-1)(x+2)}[/tex]