[tex]\displaystyle (x+1)(x^2+1)=(x+1)(4x-2) \\ \\ (x+1)(x^2+1)-(x+1)(4x-2)=0 \\ \\ (x+1)(x^2+1-4x+2)=0 \\ \\ (x+1)(x^2-4x+3)=0. \\ \\ Avem~doua~cazuri: \\ \\ i)~x+1=0 \Rightarrow x_1=-1. \\ \\ ii)~x^2-4x+3=0. \\ \\ \Delta=(-4)^2-4 \cdot 1 \cdot 3=16-12=4. \\ \\ x_2= \frac{4+\sqrt{4}}{2}= \frac{6}{2}=3. \\ \\ x_3= \frac{4- \sqrt{4}}{2}= \frac{2}{2}=1. \\ \\ Solutie:~x \in \{ -1;1;3\}.[/tex]
