Question

Determinati numerele reale a,b,c stiind ca (2x-3)³-8x+12=(x+a)(x+b)(x+c)

Answer 1

(2x-3)³-4(2x-3)=(x+a)(x+b)(x+c)
(2x-3)*[(2x-3)²-4]=(x+a)(x+b)(x+c)
(2x-3)(2x-3-4)(2x+3-4)=(x+a)(x+b)(x+c)
(2x-3)(2x-7)(2x-1)=(x+a)(x+b)(x+c)
2x-3=x+a⇒a=x-3
2x-7=x+b⇒b=x-7
2x-1=x+c⇒c=x-1
 Vf(x+x-3)(x+x-7)(x+x-1)⇒(2x-3)(2x-7)(2x-1)