Question

Exercițiul 14,vă rog.
Dau coroană.

Answer 1

   
[tex]\displaystyle\\ 14)\\ a)~3x^3+6x^2+3x=3x(x^2+2x+1)=3x(x+1)^2\\\\ b)~4x(x+1)-3(x+1)=(x+1)(4x-3)\\\\ c)\\ (x+1)^2-9=(x+1)^2-3^2=(x+1+3)(x+1-3)=(x+4)(x-2)\\\\ d)~4x^2-(x-1)^2=(2x)^2-(x-1)^2=(2x+(x-1))(2x-(x-1))=\\ =(2x+x-1)(2x-x+1))=(3x-1)(x+1)\\\\ e)~x^2+6x+8 = x^2 +2x+4x+8=\\ =x(x+2)+4(x+2)=(x+2)(x+4)\\\\ f)~x^2-9x+14=x^2-2x-7x+14=\\ =x(x-2)-7(x-2)=(x-2)(x-7)\\\\ g)~x^3+3x^2-4x-12=x^2(x+3)-4(x+3)=\\ =(x+3)(x^2-4)=(x+3)(x^2-2^2)=(x+3)(x+2)(x-2) [/tex]

[tex]\displaystyle\\ h)~x^3+2x^2-9x-18=x^2(x+2)-9(x+2)=\\ =(x+2)(x^2-9)=(x+2)(x^2-3^2)=(x+2)(x+3)(x-3)\\\\ i)~x^2-6x-16=x^2+2x-2x-6x-16=x^2+2x-8x-16=\\ =x(x+2)-8(x+2)=(x+2)(x-8)[/tex]