Question

$(x+1)(x+2)(x+3)(x+4)=15$

Answer 1

$(x+1)(x+2)(x+3)(x+4)=15\ zamena\ x+1=t\ t(t+1)(t+2)(t+3)=15\ t^4+6t^3+11t^2+6t-15=0\ t^4+3t^3+5t^2+3t^3+6t^2+6t-15=0\ t^2(t^2+3t+5)+3t^3+6t^2+6t-15=0\ t^2(t^2+3t+5)+3(t^3+6t^2+6t-15)=0\ o4evidno t=1togda\ t^2(t^2+3t+5)+3(t-1)(t^2+3t+5)=0\ (t^2+3t-3)(t^2+3t+5)=0\ t=frac{-sqrt{21}-3}{2};\ t_{2}=frac{sqrt{21}-3}{2};\ x+1=t\ x= frac{-sqrt{21}-3}{2};\ x=frac{sqrt{21}-3}{2};$